Here are the list of articles for the third Grading:
Acids and Bases
By: Mark Genesis Camba
Acids and bases are encountered frequently both in chemistry and in everyday living. They have opposite properties and have the ability to cancel or neutralize each other. Acids and bases are carefully regulated in the body by the lungs, blood, and kidneys through equilibrium processes.
What are acids and bases?
Observational definitions:
Bases:
Taste bitter.
Feels slippery
Turn red litmus paper blue.
Turn phenolphthalein pink.
Have a pH value of above 7
Bases neutralize acids producing a salt
and water
Proton acceptors
(Proton = H+ = hydrogen ion)
Increase OH- ion concentration in solution
Bases are electrolytes
Acids:
Taste sour.
Give sharp stinging pain in a cut or wound.
Turn blue litmus paper red.
Turn phenolphthalein colorless.
React with metals to produce hydrogen gas.
React with carbonates or bicarbonates to produce carbon dioxide gas.
Have a pH value of below 7
Acids neutralize bases producing a salt
and water
Proton donors
(Proton = H+ = hydrogen ion)
Increase H+ ion concentration in solution
Acids are electrolytes
Acids react with carbonates to produce a salt, water and CO2
Acids react with active metals to produce
hydrogen
Reaction of an acid with a base (neutralization) always produces a salt and water:
Acid Base Salt water
HCl (aq) + NaOH (aq) ‡ NaCl (aq) + H2O
H2SO4 (aq) + Ca(OH)2 (aq) ‡ CaSO4 (aq) + 2H2O
HNO3 (aq) + KOH (aq) ‡ KNO3 (aq) + H2O
Reaction of an acid with an active metal:
Metal Acid Salt Hydrogen gas
Zn (s) + 2HCl (aq) ‡ ZnCl2 (aq) + H2 (g)
Mg (s) + 2HNO3 (aq) ‡ Mg(NO3)2 (aq) + H2 (g)
Reaction of an acid with a carbonate:
Acid Carbonate Salt water Carbon dioxide gas
2HCl (aq) + Na2CO3 (aq) ‡ 2NaCl (aq) + H2O + CO2 (g)
H2SO4 (aq) + MgCO3 (aq) ‡ MgSO4 (aq) + H2O + CO2 (g)
By: Mark Genesis Camba
Acids and bases are encountered frequently both in chemistry and in everyday living. They have opposite properties and have the ability to cancel or neutralize each other. Acids and bases are carefully regulated in the body by the lungs, blood, and kidneys through equilibrium processes.
What are acids and bases?
Observational definitions:
Bases:
Taste bitter.
Feels slippery
Turn red litmus paper blue.
Turn phenolphthalein pink.
Have a pH value of above 7
Bases neutralize acids producing a salt
and water
Proton acceptors
(Proton = H+ = hydrogen ion)
Increase OH- ion concentration in solution
Bases are electrolytes
Acids:
Taste sour.
Give sharp stinging pain in a cut or wound.
Turn blue litmus paper red.
Turn phenolphthalein colorless.
React with metals to produce hydrogen gas.
React with carbonates or bicarbonates to produce carbon dioxide gas.
Have a pH value of below 7
Acids neutralize bases producing a salt
and water
Proton donors
(Proton = H+ = hydrogen ion)
Increase H+ ion concentration in solution
Acids are electrolytes
Acids react with carbonates to produce a salt, water and CO2
Acids react with active metals to produce
hydrogen
Reaction of an acid with a base (neutralization) always produces a salt and water:
Acid Base Salt water
HCl (aq) + NaOH (aq) ‡ NaCl (aq) + H2O
H2SO4 (aq) + Ca(OH)2 (aq) ‡ CaSO4 (aq) + 2H2O
HNO3 (aq) + KOH (aq) ‡ KNO3 (aq) + H2O
Reaction of an acid with an active metal:
Metal Acid Salt Hydrogen gas
Zn (s) + 2HCl (aq) ‡ ZnCl2 (aq) + H2 (g)
Mg (s) + 2HNO3 (aq) ‡ Mg(NO3)2 (aq) + H2 (g)
Reaction of an acid with a carbonate:
Acid Carbonate Salt water Carbon dioxide gas
2HCl (aq) + Na2CO3 (aq) ‡ 2NaCl (aq) + H2O + CO2 (g)
H2SO4 (aq) + MgCO3 (aq) ‡ MgSO4 (aq) + H2O + CO2 (g)
Back Titration (Indirect Titration)
By: Mark Geniesis G. Camba
A back titration, or indirect titration, is generally a two-stage analytical technique:
First the student pipetted 25.00 mL of the cloudy ammonia solution into a 250.0 mL conical flask.
50.00 mL of 0.100 mol L-1 HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution.
The excess (unreacted) HCl was then titrated with 0.050 mol L-1 Na2CO3(aq).
21.50 mL of Na2CO3(aq) was required.
Calculate the concentration of the ammonia in the cloudy ammonia solution.
Step 1: Determine the amount of HCl in excess from the titration results
A student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125 g sample of chalk.
The student placed the chalk sample in a 250 mL conical flask and added 50.00 mL 0.200 mol L-1 HCl using a pipette.
The excess HCl was then titrated with 0.250 mol L-1 NaOH.
The average NaOH titre was 32.12 mL
Calculate the mass of calcium carbonate, in grams, present in the chalk sample.
Step 1: Determine the amount of HCl in excess from the titration results
By: Mark Geniesis G. Camba
A back titration, or indirect titration, is generally a two-stage analytical technique:
- Reactant A of unknown concentration is reacted with excess reactant B of known concentration.
- A titration is then performed to determine the amount of reactant B in excess.
- one of the reactants is volatile, for example ammonia.
- an acid or a base is an insoluble salt, for example calcium carbonate
- a particular reaction is too slow
- direct titration would involve a weak acid - weak base titration
(the end-point of this type of direct titration is very difficult to observe)
First the student pipetted 25.00 mL of the cloudy ammonia solution into a 250.0 mL conical flask.
50.00 mL of 0.100 mol L-1 HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution.
The excess (unreacted) HCl was then titrated with 0.050 mol L-1 Na2CO3(aq).
21.50 mL of Na2CO3(aq) was required.
Calculate the concentration of the ammonia in the cloudy ammonia solution.
Step 1: Determine the amount of HCl in excess from the titration results
- Write the equation for the titration:
- Calculate the moles, n, of Na2CO3(aq) that reacted in the titration:
n = c x V
c = concentration (molarity) = 0.050 mol L-1
V = volume = 21.50 mL = 21.50 x 10-3 L
n(Na2CO3(aq)) = moles of Na2CO3(aq)= 0.050 x 21.50 x 10-3 = 1.075 x 10-3 mol - Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration.
From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl
So, 1.075 x 10-3 mole Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl
n(HCltitrated) = 2 x 1.075 x 10-3 = 2.150 x 10-3 mol - The amount of HCl that was added to the cloudy ammonia solution in excess was
2.150 x 10-3 mol
- Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution:
n(HCltotal added) = c x V
c = concentration (molarity) = 0.100 mol L-1
V = volume = 50.00 mL = 50.00 x 10-3 L
n(HCltotal added) = 0.100 x 50.00 x 10-3 = 5.00 x 10-3 mol - Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution
n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added)
n(HCltotal added) = 5.00 x 10-3 mol
n(HCltitrated) = 2.150 x 10-3 mol
2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3
n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 mol - Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).
- From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl.
From the equation, 1 mol HCl reacts with 1 mol NH3
So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution. - Calculate the ammonia concentration in the cloudy ammonia solution.
c = n ÷ V
n = moles (NH3) = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl)
V = volume (NH3(aq)) = 25.00 mL = 25.00 x 10-3 L (volume of ammonia solution that reacted with HCl)
c = concentration (molarity) = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 mol L-1 - The concentration of ammonia in the cloudy ammonia solution was 0.114 mol L-1
A student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125 g sample of chalk.
The student placed the chalk sample in a 250 mL conical flask and added 50.00 mL 0.200 mol L-1 HCl using a pipette.
The excess HCl was then titrated with 0.250 mol L-1 NaOH.
The average NaOH titre was 32.12 mL
Calculate the mass of calcium carbonate, in grams, present in the chalk sample.
Step 1: Determine the amount of HCl in excess from the titration results
- Write the equation for the titration:
- Calculate the moles, n, of NaOH(aq) that reacted in the titration:
n = c x V
c = concentration = 0.250 mol L-1
V = volume = 32.12 mL = 32.12 x 10-3 L
n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol - Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration.
From the balanced chemical equation, 1 mole NaOH reacts with 1 mole of HCl
So, 8.03 x 10-3 mole NaOH reacted with 8.03 x 10-3 moles HCl - The amount of HCl that was added to the chalk in excess was
8.03 x 10-3 mol
- Calculate the total moles of HCl, n(HCl), originally added to the chalk:
n(HCltotal added) = c x V
c = concentration (molarity) = 0.200 mol L-1
V = volume = 50.00 mL = 50.00 x 10-3 L
n(HCltotal added) = 0.200 x 50.00 x 10-3 = 0.010 mol - Calculate the moles of HCl that reacted with the calcium carbonate in the chalk
n(HCltitrated) + n(HClreacted with calcium carbonate) = n(HCltotal added)
n(HCltotal added) = 0.010 mol
n(HCltitrated) = 8.03 x 10-3 mol
8.03 x 10-3 + n(HClreacted with calcium carbonate) = 0.010
n(HClreacted with calcium carbonate) = 0.010 - 8.03 x 10-3 = 1.97 x 10-3 mol - Write the balanced chemical equation for the reaction between calcium carbonate, CaCO3(s), in the chalk and the HCl(aq).
- From the balanced chemical equation, calculate the moles of CaCO3 that reacted with HCl.
From the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1 mol HCl reacts with ½ mol CaCO3
So, 1.97 x 10-3 mol HCl had reacted with ½ x 1.97 x 10-3 = 9.85 x 10-4 mol CaCO3 in the chalk. - Calculate the mass of calcium carbonate in the chalk.
moles = mass ÷ molar mass
moles = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl)
molar mass(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol-1
mass = moles x molar mass = 9.85 x 10-4 x 100.09 = 0.099 g - The mass of calcium carbonate in the chalk was 0.099 g
STANDARDIZATION
By: Maricon Ivee Barlis
Standardization is doing a titration to work out the exact concentration of the solution you want to use to determine the concentration of an unknown solution.
To do a titration you must to know the exact number of moles of one of the reagents (titrant) you are using so that you can then determine the number of moles of the unknown reagent.
There are some commonly used titrants that absorb water, so when you weigh them out you are weighing water as well, so you can never be quite sure exactly how many moles you weigh out.
Some titrants do not stay the same concentration that you make them up at either. They may absorb water from the air, or decompose a little. So when you go to use it a week later you may not know it's exact concentration.
For these types of reagents it is common to "standardize" them before you use them.
It basically means you do a series of 2 or 3 titrations with the reagent you want to use against a third reagent that you definitely know the concentration of. These are called primary standards, they are typically compounds that you can weigh out exact amounts of and be sure that that is exactly what you get.
Since you know the concentration of the primary reagent you can use the titration to determine the exact concentration of the titrant that you want to use. Then, happy in the knowledge that you have accurately determined the concentration ("standardized") your reagent, you can then go ahead and use it to determine the concentration of your unknown.
ACID-BASE TITRATION
In an acid-base titration the idea is to carry out a neutralization reaction between an acid solution and a base solution in order to determine the concentration of either an unknown acid or base solution. This requires that you know the concentration of the other solution as accurately as possible.
Determining the precise concentration of a solution just by volumes is not very accurate due to the error involved in reading the scales. It is far more accurate to weigh a solid material with a scale and then use that accurately known quantity to determine the concentration of the test solution. The test solution is then used to carry out the acid-base titration.
For example, you could accurately weigh out 40 g of NaOH and dissolve it in a little water, then add enough water to fill a 1L volumetric flask. This solution would then be used to titrate an accurately weighed amount of a crystalline solid acid, or reference compound. This will give you a very accurate determination of the concentration of the NaOH solution. This is how the NaOH solution has been standardized, and can subsequently be used to accurately determine the concentration of an acid solution.
EXAMPLE:
An aqueous solution of sodium hydroxide (NaOH) was standardized by titrating it against
a 0.1421 g sample of potassium hydrogen phthalate (KHP). (The chemical formula of KHP is HKC8H4O4.)
The initial reading taken from the buret containing the NaOH was 0.52 mL and the final volume reading
was 32.86 mL.
What is the molarity of the NaOH solution?
The reaction equation is:
HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O
Volume of NaOH used:
32.86 mL – 0.52 mL = 32.34 mL NaOH
Moles of KHP (HKC8H4O4):
(0.1421 g KHP)(1 mol KHP / 204.23 g KHP ) = 6.958 x 10-4 mol KHP
Moles of NaOH neutralized:
(6.958 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.958 x 10-4 mol NaOH
Molarity of NaOH:
6.958 x 10-4 mol NaOH / 0.03234 L NaOH = 0.02152 M NaOH
By: Maricon Ivee Barlis
Standardization is doing a titration to work out the exact concentration of the solution you want to use to determine the concentration of an unknown solution.
To do a titration you must to know the exact number of moles of one of the reagents (titrant) you are using so that you can then determine the number of moles of the unknown reagent.
There are some commonly used titrants that absorb water, so when you weigh them out you are weighing water as well, so you can never be quite sure exactly how many moles you weigh out.
Some titrants do not stay the same concentration that you make them up at either. They may absorb water from the air, or decompose a little. So when you go to use it a week later you may not know it's exact concentration.
For these types of reagents it is common to "standardize" them before you use them.
It basically means you do a series of 2 or 3 titrations with the reagent you want to use against a third reagent that you definitely know the concentration of. These are called primary standards, they are typically compounds that you can weigh out exact amounts of and be sure that that is exactly what you get.
Since you know the concentration of the primary reagent you can use the titration to determine the exact concentration of the titrant that you want to use. Then, happy in the knowledge that you have accurately determined the concentration ("standardized") your reagent, you can then go ahead and use it to determine the concentration of your unknown.
ACID-BASE TITRATION
In an acid-base titration the idea is to carry out a neutralization reaction between an acid solution and a base solution in order to determine the concentration of either an unknown acid or base solution. This requires that you know the concentration of the other solution as accurately as possible.
Determining the precise concentration of a solution just by volumes is not very accurate due to the error involved in reading the scales. It is far more accurate to weigh a solid material with a scale and then use that accurately known quantity to determine the concentration of the test solution. The test solution is then used to carry out the acid-base titration.
For example, you could accurately weigh out 40 g of NaOH and dissolve it in a little water, then add enough water to fill a 1L volumetric flask. This solution would then be used to titrate an accurately weighed amount of a crystalline solid acid, or reference compound. This will give you a very accurate determination of the concentration of the NaOH solution. This is how the NaOH solution has been standardized, and can subsequently be used to accurately determine the concentration of an acid solution.
EXAMPLE:
An aqueous solution of sodium hydroxide (NaOH) was standardized by titrating it against
a 0.1421 g sample of potassium hydrogen phthalate (KHP). (The chemical formula of KHP is HKC8H4O4.)
The initial reading taken from the buret containing the NaOH was 0.52 mL and the final volume reading
was 32.86 mL.
What is the molarity of the NaOH solution?
The reaction equation is:
HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O
Volume of NaOH used:
32.86 mL – 0.52 mL = 32.34 mL NaOH
Moles of KHP (HKC8H4O4):
(0.1421 g KHP)(1 mol KHP / 204.23 g KHP ) = 6.958 x 10-4 mol KHP
Moles of NaOH neutralized:
(6.958 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.958 x 10-4 mol NaOH
Molarity of NaOH:
6.958 x 10-4 mol NaOH / 0.03234 L NaOH = 0.02152 M NaOH
Acid-Base Definition
By: Nichole T. Santos
This page describes the Arrhenius, Bronsted-Lowry, and Lewis theories of acids and bases, and explains the relationships between them. It also explains the concept of a conjugate pair- an acid and its conjugate base, or a base and it’s conjugate acid.
The Arrhenius Theory of acids and bases
- Acids are substances which produce hydrogen ions in solution.
- Bases are substances which produce hydroxide ions in solution
Neutralisation happens because hydrogen ions and hydroxide ions react to produce water.
H+(aq) + OH-(aq) --------à H2O(l)
Limitations of the Theory
Hydrochloric acid is neutralized by both sodium hydroxide solution and ammonia solution. In both cases, you get a colourless solution which you can crstallise to get a white salt- either sodium chloride or ammonium chloride.
There are clearly very similar reactions. The full equations are:
NaOH(aq) + HCL(aq) ------à NaCl(aq) + H2O(l)
NH3(aq) + HCL(aq) --------à NH4+Cl(aq)
In the sodium hydroxide case, hydrogen ions from the acid are reacting with hydroxide ions from the sodium hydroxide – in line with Arrhenius theory.
However, in the ammonia case, there don’t appear to be any hydroxide ions:
This is a reversible reaction, and in a typical dilute ammonia solution, about 99% of the ammonia remains as ammonia molecules. Nevertheless, there are hydroxide ions there and we can squeeze this into the Arrhenius theory.
However, this same reaction also happens between ammonia gas and hydrogen chloride gas.
NH3(g) + HCl(g) -------à NH4Cl(s)
In this case, there aren’t any hydrogen ions or hydroxide ions in solution – because there isn’t any solution. The Arrhenius theory wouldn’t count this as an acid-base reaction, despite the fact that it is producing the same product as when the two substances were in solution.
The Bronsted-Lowry Theory of acids and bases
- An acid is a proton (hydrogen ion) donor.
- A base is a proton (hydrogen ion) acceptor.
The relationship between the Bronsted-Lowry theory and the Arrhenius theory
The Brownsted-Lowry theory doesn’t go against the Arrhenius theory in any way – it just adds to it.
Hydroxide ions are still bases because they accept hydrogen ions from acids and form water.
An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.
When hydrogen chloride gas dissolves in water to produce hydrochloric acid, the hydrogen chloride molecule gives a proton (a hydrogen ion) to a water molecule. A co-ordinate (dative covalent) cond is formed between one of the lone pairs on the oxygen and the hydrogen from the HCl. Hydroxoniums ions, H3O+ are produced.
H2O + HCl --------à H3O+ + Cl-
When an acid in solution reacts with a bvase, what is actually functioning as the acid is the hydroxonium ion. For example, a proton is transferred from a hydroxonium ion to a hydroxide ion to make water.
H3O+(aq) + OH- --------à 2H2O(l)
Showing the electrons, but leaving out the inner ones:
It is important to realize the whenever you talk about hydrogen ions in solution, H+ (aq), what you are actually talking about are hydroxonium ions.
The hydrogen chloride/ammonia problem
This is no longer a problem using the Brownsted-Lowry theory. Whether you are talking about the reaction in solution or in the gas state, ammonia is a base beacause it accepts a proton (a hydrogen ion). The hydrogen becomes attached to the lone pair on the nitrogen of the ammonia via a co-ordinate bond.
It is in solution, the ammonia accepts a proton from a hydroxonium ion:
NH3(aq) + H3O+(aq) ----------à NH4+(s) + H2O(l)
If the reaction is happening in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
NH3(aq) + HCl(g) ----------à NH4+(s) + Cl-(s)
Either way, the ammonia acts as a base by accepting a hydrogen ion from an acid.
Conjugate pairs
When hydrogen chloride dissolves in water, almost 100% of it reacts with the water to produce hydroxonium ions and chloride ions. Hydrogen chloride is a strong acid, and we tend to write this as a one-way reaction:
H2O + HCl ------à H3O+ + Cl-
In fact, the reaction between HCl and water is reversible, but only to a very minor extent. In order to generalize, consider an acid HA, and think of the reaction as being reversible.
Thinking avout the forward reaction:
-The Ha is an acid because it is donating a proton (hydrogen ion) to the water.
- The water is a base because it is accepting a proton from the Ha.
But there is also a back reaction between the hydroxonium ion and the A- ion:
- The H3O+ is an acid because it is donating a proton (hydrogen ion) to the A- ion.
- The A- ion is a base because it is accepting a proton from the H3O+
The reversible reaction contains two acids and two bases. We think of them in pairs, called conjugate pairs.
When the acid, HA, loses a proton it forms a base, A-. When the base, A-,
Accepts a proton back again, it obviously refoms the acid, Ha. These two are a conjugate pair.
The Lewis Theory of acids and bases
-an acid is an electron pair acceptor.
- A base is an electron pair donor.
The relationship between the Lewis theory and the Bronsted-Lowry theory
Lewis bases
It is easiest to see the relationship by looking at exactly what Bronsted-Lowry bases do when they accept hydrogen ions. Three Bronsted-Lowry bases we’ve lloked at are hydroxide ions, ammonia and water, and they are typical of all the rest.
The Bronsted-Lowry theory says that they are acting as bases because they are combining with hydrogen ions. The reason they are combining with hyfrogen ions is that they have lone pair of electrons – which is what the Lewis theory says. The two are entirely consistent.
As far as the ammonia is concerned, it it behaving exactly the same as when it reacts with a hydrogen ion – it is using its lone pair to from a co-ordinate bond. If you are going to describe it as a base in one case, it makes sense to describe it as one in the other cases as well.
Lewis acids
- A lewis acis is an electron pair acceptor.
- A lewis base is an electron pair donor
In the above example, the BF3 is acting as the lewis acid by accepting the nitrogen’s lone pair. On the Bronsted-Lowry theory, the BF3 has nothing remotely acidic about it
This is an extension of the term acid well beyong any commom use.
By: Nichole T. Santos
This page describes the Arrhenius, Bronsted-Lowry, and Lewis theories of acids and bases, and explains the relationships between them. It also explains the concept of a conjugate pair- an acid and its conjugate base, or a base and it’s conjugate acid.
The Arrhenius Theory of acids and bases
- Acids are substances which produce hydrogen ions in solution.
- Bases are substances which produce hydroxide ions in solution
Neutralisation happens because hydrogen ions and hydroxide ions react to produce water.
H+(aq) + OH-(aq) --------à H2O(l)
Limitations of the Theory
Hydrochloric acid is neutralized by both sodium hydroxide solution and ammonia solution. In both cases, you get a colourless solution which you can crstallise to get a white salt- either sodium chloride or ammonium chloride.
There are clearly very similar reactions. The full equations are:
NaOH(aq) + HCL(aq) ------à NaCl(aq) + H2O(l)
NH3(aq) + HCL(aq) --------à NH4+Cl(aq)
In the sodium hydroxide case, hydrogen ions from the acid are reacting with hydroxide ions from the sodium hydroxide – in line with Arrhenius theory.
However, in the ammonia case, there don’t appear to be any hydroxide ions:
This is a reversible reaction, and in a typical dilute ammonia solution, about 99% of the ammonia remains as ammonia molecules. Nevertheless, there are hydroxide ions there and we can squeeze this into the Arrhenius theory.
However, this same reaction also happens between ammonia gas and hydrogen chloride gas.
NH3(g) + HCl(g) -------à NH4Cl(s)
In this case, there aren’t any hydrogen ions or hydroxide ions in solution – because there isn’t any solution. The Arrhenius theory wouldn’t count this as an acid-base reaction, despite the fact that it is producing the same product as when the two substances were in solution.
The Bronsted-Lowry Theory of acids and bases
- An acid is a proton (hydrogen ion) donor.
- A base is a proton (hydrogen ion) acceptor.
The relationship between the Bronsted-Lowry theory and the Arrhenius theory
The Brownsted-Lowry theory doesn’t go against the Arrhenius theory in any way – it just adds to it.
Hydroxide ions are still bases because they accept hydrogen ions from acids and form water.
An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.
When hydrogen chloride gas dissolves in water to produce hydrochloric acid, the hydrogen chloride molecule gives a proton (a hydrogen ion) to a water molecule. A co-ordinate (dative covalent) cond is formed between one of the lone pairs on the oxygen and the hydrogen from the HCl. Hydroxoniums ions, H3O+ are produced.
H2O + HCl --------à H3O+ + Cl-
When an acid in solution reacts with a bvase, what is actually functioning as the acid is the hydroxonium ion. For example, a proton is transferred from a hydroxonium ion to a hydroxide ion to make water.
H3O+(aq) + OH- --------à 2H2O(l)
Showing the electrons, but leaving out the inner ones:
It is important to realize the whenever you talk about hydrogen ions in solution, H+ (aq), what you are actually talking about are hydroxonium ions.
The hydrogen chloride/ammonia problem
This is no longer a problem using the Brownsted-Lowry theory. Whether you are talking about the reaction in solution or in the gas state, ammonia is a base beacause it accepts a proton (a hydrogen ion). The hydrogen becomes attached to the lone pair on the nitrogen of the ammonia via a co-ordinate bond.
It is in solution, the ammonia accepts a proton from a hydroxonium ion:
NH3(aq) + H3O+(aq) ----------à NH4+(s) + H2O(l)
If the reaction is happening in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
NH3(aq) + HCl(g) ----------à NH4+(s) + Cl-(s)
Either way, the ammonia acts as a base by accepting a hydrogen ion from an acid.
Conjugate pairs
When hydrogen chloride dissolves in water, almost 100% of it reacts with the water to produce hydroxonium ions and chloride ions. Hydrogen chloride is a strong acid, and we tend to write this as a one-way reaction:
H2O + HCl ------à H3O+ + Cl-
In fact, the reaction between HCl and water is reversible, but only to a very minor extent. In order to generalize, consider an acid HA, and think of the reaction as being reversible.
Thinking avout the forward reaction:
-The Ha is an acid because it is donating a proton (hydrogen ion) to the water.
- The water is a base because it is accepting a proton from the Ha.
But there is also a back reaction between the hydroxonium ion and the A- ion:
- The H3O+ is an acid because it is donating a proton (hydrogen ion) to the A- ion.
- The A- ion is a base because it is accepting a proton from the H3O+
The reversible reaction contains two acids and two bases. We think of them in pairs, called conjugate pairs.
When the acid, HA, loses a proton it forms a base, A-. When the base, A-,
Accepts a proton back again, it obviously refoms the acid, Ha. These two are a conjugate pair.
The Lewis Theory of acids and bases
-an acid is an electron pair acceptor.
- A base is an electron pair donor.
The relationship between the Lewis theory and the Bronsted-Lowry theory
Lewis bases
It is easiest to see the relationship by looking at exactly what Bronsted-Lowry bases do when they accept hydrogen ions. Three Bronsted-Lowry bases we’ve lloked at are hydroxide ions, ammonia and water, and they are typical of all the rest.
The Bronsted-Lowry theory says that they are acting as bases because they are combining with hydrogen ions. The reason they are combining with hyfrogen ions is that they have lone pair of electrons – which is what the Lewis theory says. The two are entirely consistent.
As far as the ammonia is concerned, it it behaving exactly the same as when it reacts with a hydrogen ion – it is using its lone pair to from a co-ordinate bond. If you are going to describe it as a base in one case, it makes sense to describe it as one in the other cases as well.
Lewis acids
- A lewis acis is an electron pair acceptor.
- A lewis base is an electron pair donor
In the above example, the BF3 is acting as the lewis acid by accepting the nitrogen’s lone pair. On the Bronsted-Lowry theory, the BF3 has nothing remotely acidic about it
This is an extension of the term acid well beyong any commom use.
Acid-Base Indicator
By: Melissa Bugayon
What is an acid-base indicator?
An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.
How is an indicator used?
Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.
What are some common acid-base indicators?
Several acid-base indicators are listed below, some more than once if they can be used over multiple pH ranges. Quantity of indicator in aqueous (aq.) or alcohol (alc.) solution is specified. Tried-and-true indicators include: thymol blue, tropeolin OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, phenolphthalein, thymolphthalein, alizarin yellow, tropeolin O, nitramine, and trinitrobenzoic acid. Data in this table are for sodium salts of thymol blue, bromphenol blue, tetrabromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, and cresol red.
By: Melissa Bugayon
What is an acid-base indicator?
An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.
How is an indicator used?
Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.
What are some common acid-base indicators?
Several acid-base indicators are listed below, some more than once if they can be used over multiple pH ranges. Quantity of indicator in aqueous (aq.) or alcohol (alc.) solution is specified. Tried-and-true indicators include: thymol blue, tropeolin OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, phenolphthalein, thymolphthalein, alizarin yellow, tropeolin O, nitramine, and trinitrobenzoic acid. Data in this table are for sodium salts of thymol blue, bromphenol blue, tetrabromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, and cresol red.
Titration
By: Michael Angelo T. Tablante
The word titration comes from the Latin word "titulus", which means inscription or title. The French word titre means rank. Therefore, Titration means the determination of concentration or Rank of a solution with respect to water with a pH of 7.
The standard solution is usually added from a graduated vessel called a burette. The process of adding standard solution until the reaction is just complete is termed as titration and the substance to be determined is said to be titrated.
All chemical reactions cannot be considered as titrations. A reaction can serve as a basis of a titration procedure only if the following conditions are satisfied:
1. The reaction must be a fast one.
2. It must proceed stoichiometrically.
3. The change in free energy (ΔG) during the reaction must be sufficiently large for spontaneity of the reaction.
4. There should be a way to detect the completion of the reaction.
End point and Equivalent point:
For a reaction, a stage which shows the completion of a particular reaction is known as end point. Equivalence point is a stage in which the amount of reagent added is exactly and stoichiometrically equivalent to the amount of the reacting substance in the titrated solution. The end point is detected by some physical change produced by the solution, by itself or more usually by the addition of an auxiliary reagent known as an 'indicator'. The end point and the equivalence point may not be identical. End point is usually detected only after adding a slight excess of the titrant. In many cases, the difference between these two will fall within the experimental error.
Indicator:
It is a chemical reagent used to recognize the attainment of end point in a titration. After the reaction between the substance and the standard solution is complete, the indicator should give a clear colour change.
When a titration is carried out, the free energy change for the reaction is always negative.
That is, during the initial stages of the reaction between A & B, when the titrant A is added to B the following reaction takes place.
Equilibrium constant,
a = activity co-efficient.
Large values of the equilibrium constant K implies that the equilibrium concentration of A & B are very small at the equivalence point. It also indicates that the reverse reaction is negligible and the product C & D are very much more stable than the reactants A & B. Greater the value of K the larger the magnitude of the negative free energy change for the reaction between A & B. Since,
Where,
R = Universal gas Constant = 8.314 JK-1mol-1,
T = Absolute Temperature.
The reaction of the concentration of A & B leads to the reduction of the total free energy change. If the concentrations of A & B are too low the magnitude of the total free energy change becomes so small and the use of the reaction for titration will not be feasible.
Expressions of Concentration of Solutions:
The concentration or strength of solution means the amount of solute present in a given amount of the solution. The concentration may be expressed in physical or chemical units.
1. Normality (N): It is defined as number of gram equivalents of the solute present in 1 litre (1000mL.) of the solution. If W g of solute of equivalent weight E is present in V mL of the solution, the normality of the solution is given by:
2. Molarity (M): It is defined as the number of moles of the solute present in 1 litre (or 1000 mL) of the solution. A one molar solution contains 1 mole of the solute dissolved in 1 litre of the solution.
3. Molality (m): It is defined as the number of moles of solute dissolved in 1000 g of the solvent. One molal solution contains one mole of the solute dissolved in 1000 g of the solvent.
By: Michael Angelo T. Tablante
The word titration comes from the Latin word "titulus", which means inscription or title. The French word titre means rank. Therefore, Titration means the determination of concentration or Rank of a solution with respect to water with a pH of 7.
The standard solution is usually added from a graduated vessel called a burette. The process of adding standard solution until the reaction is just complete is termed as titration and the substance to be determined is said to be titrated.
All chemical reactions cannot be considered as titrations. A reaction can serve as a basis of a titration procedure only if the following conditions are satisfied:
1. The reaction must be a fast one.
2. It must proceed stoichiometrically.
3. The change in free energy (ΔG) during the reaction must be sufficiently large for spontaneity of the reaction.
4. There should be a way to detect the completion of the reaction.
End point and Equivalent point:
For a reaction, a stage which shows the completion of a particular reaction is known as end point. Equivalence point is a stage in which the amount of reagent added is exactly and stoichiometrically equivalent to the amount of the reacting substance in the titrated solution. The end point is detected by some physical change produced by the solution, by itself or more usually by the addition of an auxiliary reagent known as an 'indicator'. The end point and the equivalence point may not be identical. End point is usually detected only after adding a slight excess of the titrant. In many cases, the difference between these two will fall within the experimental error.
Indicator:
It is a chemical reagent used to recognize the attainment of end point in a titration. After the reaction between the substance and the standard solution is complete, the indicator should give a clear colour change.
When a titration is carried out, the free energy change for the reaction is always negative.
That is, during the initial stages of the reaction between A & B, when the titrant A is added to B the following reaction takes place.
Equilibrium constant,
a = activity co-efficient.
Large values of the equilibrium constant K implies that the equilibrium concentration of A & B are very small at the equivalence point. It also indicates that the reverse reaction is negligible and the product C & D are very much more stable than the reactants A & B. Greater the value of K the larger the magnitude of the negative free energy change for the reaction between A & B. Since,
Where,
R = Universal gas Constant = 8.314 JK-1mol-1,
T = Absolute Temperature.
The reaction of the concentration of A & B leads to the reduction of the total free energy change. If the concentrations of A & B are too low the magnitude of the total free energy change becomes so small and the use of the reaction for titration will not be feasible.
Expressions of Concentration of Solutions:
The concentration or strength of solution means the amount of solute present in a given amount of the solution. The concentration may be expressed in physical or chemical units.
1. Normality (N): It is defined as number of gram equivalents of the solute present in 1 litre (1000mL.) of the solution. If W g of solute of equivalent weight E is present in V mL of the solution, the normality of the solution is given by:
2. Molarity (M): It is defined as the number of moles of the solute present in 1 litre (or 1000 mL) of the solution. A one molar solution contains 1 mole of the solute dissolved in 1 litre of the solution.
3. Molality (m): It is defined as the number of moles of solute dissolved in 1000 g of the solvent. One molal solution contains one mole of the solute dissolved in 1000 g of the solvent.
Polyprotic Acids And Base
By: Renz Christian V. Marcelo
Polyprotic acids are specific acids that are capable of losing more than a single proton per molecule in acid-base reactions. (In other words, acids that have more than one ionizable H+ atom per molecule). Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. Contrast with monoprotic acids in section Monoprotic Versus Polyprotic Acids And Bases.
Polyprotic Bases are bases that can accept at least one H+ ion, or proton, in acid-base reactions.
Sample Problem
What is the pH of the resulting solution when 50.0 mL of 0.700 M NaOH is mixed with 50.0 mL of 0.500 M H3PO4?
1. It is easiest to consider the initial reactions stepwise:
moles of H3PO4 = (0.0500 L)(0.500 M) = 0.0250 moles
moles of OH- = (0.0500 L)(0.700 M) = 0.0350 moles
0.0250 moles of H3PO4 neutralizes 0.0250 moles of OH-. We now have 0.0100 moles of OH- and 0.0250 moles of H2PO4-. The original amount of H3PO4 is completely consumed.
0.0100 moles of H2PO4- neutralizes the remaining OH-, leaving 0.0150 moles of H2PO4- and 0.0100 moles of HPO42-.
2. So the equilibrium is: H2PO4-(aq) HPO42-(aq) + H+(aq) Ka2 = 6.34x10-8
and the equilibrium expression is:
[H+][HPO42-]
Ka2 = ------------
[H2PO4-]
3. and 4. We have formed a buffer so steps 3. and 4. are not needed explicitly.
5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:
[H+][HPO42-]
Ka2 = ------------
[H2PO4-]
[H+](0.0100 moles/0.100 L)
6.34x10-8 = ------------------------
(0.0150 moles/0.100 L)
Note that the volume appears in the numerator and denominator and cancels out.
[H+] = (6.34x10-8) * (0.0150) / (0.0100)
[H+] = 9.51x10-8
This problem asked for the pH of the solution.
pH = -log[H+]
pH = -log(9.51x10-8)
pH=7.02
By: Renz Christian V. Marcelo
Polyprotic acids are specific acids that are capable of losing more than a single proton per molecule in acid-base reactions. (In other words, acids that have more than one ionizable H+ atom per molecule). Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. Contrast with monoprotic acids in section Monoprotic Versus Polyprotic Acids And Bases.
Polyprotic Bases are bases that can accept at least one H+ ion, or proton, in acid-base reactions.
Sample Problem
What is the pH of the resulting solution when 50.0 mL of 0.700 M NaOH is mixed with 50.0 mL of 0.500 M H3PO4?
1. It is easiest to consider the initial reactions stepwise:
moles of H3PO4 = (0.0500 L)(0.500 M) = 0.0250 moles
moles of OH- = (0.0500 L)(0.700 M) = 0.0350 moles
0.0250 moles of H3PO4 neutralizes 0.0250 moles of OH-. We now have 0.0100 moles of OH- and 0.0250 moles of H2PO4-. The original amount of H3PO4 is completely consumed.
0.0100 moles of H2PO4- neutralizes the remaining OH-, leaving 0.0150 moles of H2PO4- and 0.0100 moles of HPO42-.
2. So the equilibrium is: H2PO4-(aq) HPO42-(aq) + H+(aq) Ka2 = 6.34x10-8
and the equilibrium expression is:
[H+][HPO42-]
Ka2 = ------------
[H2PO4-]
3. and 4. We have formed a buffer so steps 3. and 4. are not needed explicitly.
5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:
[H+][HPO42-]
Ka2 = ------------
[H2PO4-]
[H+](0.0100 moles/0.100 L)
6.34x10-8 = ------------------------
(0.0150 moles/0.100 L)
Note that the volume appears in the numerator and denominator and cancels out.
[H+] = (6.34x10-8) * (0.0150) / (0.0100)
[H+] = 9.51x10-8
This problem asked for the pH of the solution.
pH = -log[H+]
pH = -log(9.51x10-8)
pH=7.02
Diprotic Acids
By: Daryl Dave Crisostomo
A diprotic acid is an acid that can donate two proton or hydrogen atom per molecule to an aqueous solution.
The acid equilibrium problems discussed so far have focused on a family of compounds known as monoprotic acids. Each of these acids has a single H+ ion, or proton, it can donate when it acts as a Brnsted acid. Hydrochloric acid (HCl), acetic acid (CH3CO2H or HOAc), nitric acid (HNO3), and benzoic acid (C6H5CO2H) are all monoprotic acids.
Several important acids can be classified as polyprotic acids, which can lose more than one H+ ion when they act as Brnsted acids. Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4) have two acidic hydrogen atoms. Triprotic acids, such as phosphoric acid (H3PO4) and citric acid (C6H8O7), have three.
There is usually a large difference in the ease with which these acids lose the first and second (or second and third) protons. When sulfuric acid is classified as a strong acid, students often assume that it loses both of its protons when it reacts with water. That isn't a legitimate assumption. Sulfuric acid is a strong acid because Ka for the loss of the first proton is much larger than 1. We therefore assume that essentially all the H2SO4 molecules in an aqueous solution lose the first proton to form the HSO4-, or hydrogen sulfate, ion.
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)
Ka1 = 1 x 103
But Ka for the loss of the second proton is only 10-2 and only 10% of the H2SO4 molecules in a 1 Msolution lose a second proton.
HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq)
Ka2 = 1.2 x 10-2
H2SO4 only loses both H+ ions when it reacts with a base, such as ammonia.
The table below gives values of Ka for some common polyprotic acids. The large difference between the values of Ka for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a timean assumption known as stepwise dissociation.
Acid-Dissociation Equilibrium Constants for Common Polyprotic Acids
Acid
Ka1
Ka2
Ka3
sulfuric acid (H2SO4)
1.0 x 103
1.2 x 10-2
chromic acid (H2CrO4)
9.6
3.2 x 10-7
oxalic acid (H2C2O4)
5.4 x 10-2
5.4 x 10-5
sulfurous acid (H2SO3)
1.7 x 10-2
6.4 x 10-8
phosphoric acid (H3PO4)
7.1 x 10-3
6.3 x 10-8
4.2 x 10-13
glycine (C2H6NO2)
4.5 x 10-3
2.5 x 10-10
citric acid (C6H8O7)
7.5 x 10-4
1.7 x 10-5
4.0 x 10-7
carbonic acid (H2CO3)
4.5 x 10-7
4.7 x 10-11
hydrogen sulfide (H2S)
1.0 x 10-7
1.3 x 10-13
Let's look at the consequence of the assumption that polyprotic acids lose protons one step at a time by examining the chemistry of a saturated solution of H2S in water.
Hydrogen sulfide is the foul-smelling gas that gives rotten eggs their unpleasant odor. It is an excellent source of the S2- ion, however, and is therefore commonly used in introductory chemistry laboratories. H2S is a weak acid that dissociates in steps. Some of the H2S molecules lose a proton in the first step to form the HS-, or hydrogen sulfide, ion.
First step:
H2S(aq) + H2O(l) H3O+(aq) + HS-(aq)
A small fraction of the HS- ions formed in this reaction then go on to lose another H+ ion in a second step.
Second step:
HS-(aq) + H2O(l) H3O+(aq) + S2-(aq)
Since there are two steps in this reaction, we can write two equilibrium constant expressions.
Although each of these equations contains three terms, there are only four unknowns[H3O+], [H2S], [HS-], and [S2-] because the [H3O+] and [HS-] terms appear in both equations. The [H3O+] term represents the total H3O+ ion concentration from both steps and therefore must have the same value in both equations. Similarly, the [HS-] term, which represents the balance between the HS- ions formed in the first step and the HS- ions consumed in the second step, must have the same value for both equations.
Four equations are needed to solve for four unknowns. We already have two equations: the Ka1 andKa2 expressions. We are going to have to find either two more equations or a pair of assumptions that can generate two equations. We can base one assumption on the fact that the value of Ka1 for this acid is almost a million times larger than the value of Ka2.
Ka1 >> Ka2
This means that only a small fraction of the HS- ions formed in the first step go on to dissociate in the second step. If this is true, most of the H3O+ ions in this solution come from the dissociation of H2S, and most of the HS- ions formed in this reaction PSS remain in solution. As a result, we can assume that the H3O+ and HS- ion concentrations are more or less equal.
First assumption:
[H3O+][HS-]
We need one more equation, and therefore one more assumption. Note that H2S is a weak acid (Ka1= 1.0 x 10-7, Ka2 = 1.3 x 10-13). Thus, we can assume that most of the H2S that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H2S is approximately equal to the initial concentration.
Second assumption:
[H2S] CH2S
We now have four equations in four unknowns.
[H3O+] [HS-]
[H2S] CH2S
Since there is always a unique solution to four equations in four unknowns, we are now ready to calculate the H3O+, H2S, HS-, and S2- concentrations at equilibrium in a saturated solution of H2S in water. All we need to know is that a saturated solution of H2S in water has an initial concentration of about 0.10 M.
Because Ka1 is so much larger than Ka2 for this acid, we can work with the equilibrium expression for the first step without worrying about the second step for the moment. We therefore start with the expression for Ka1 for this acid.
We then invoke one of our assumptions.
[H2S] CH2S 0.10 M
Substituting this approximation into the Ka1 expression gives the following equation.
We then invoke the other assumption.
[H3O+] [HS-]C
Substituting this approximation into the Ka1 expression gives the following result.
We now solve this approximate equation for C.
C 1.0 x 10-4
If our two assumptions are valid, we are three-fourths of the way to our goal. We know the H2S, H3O+, and HS- concentrations.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
Having extracted the values of three unknowns from the first equilibrium expression, we turn to the second equilibrium expression.
Substituting the known values of the H3O+ and HS- ion concentrations into this expression gives the following equation.
Because the equilibrium concentrations of the H3O+ and HS- ions are more or less the same, the S2-ion concentration at equilibrium is approximately equal to the value of Ka2 for this acid.
[S2-] 1.3 x 10-13 M
It is now time to check our assumptions. Is the dissociation of H2S small compared with the initial concentration? Yes. The HS- and H3O+ ion concentrations obtained from this calculation are 1.0 x 10-4 M, which is 0.1% of the initial concentration of H2S. The following assumption is therefore valid.
[H2S] CH2S 0.10 M
Is the difference between the S2- and HS- ion concentrations large enough to allow us to assume that essentially all of the H3O+ ions at equilibrium are formed in the first step and that essentially all of the HS- ions formed in this step remain in solution? Yes. The S2- ion concentration obtained from this calculation is 109 times smaller than the HS- ion concentration. Thus, our other assumption is also valid.
[H3O+] [HS-]
We can therefore summarize the concentrations of the various components of this equilibrium as follows.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
[S2-] 1.3 x 10-13 M
By: Daryl Dave Crisostomo
A diprotic acid is an acid that can donate two proton or hydrogen atom per molecule to an aqueous solution.
The acid equilibrium problems discussed so far have focused on a family of compounds known as monoprotic acids. Each of these acids has a single H+ ion, or proton, it can donate when it acts as a Brnsted acid. Hydrochloric acid (HCl), acetic acid (CH3CO2H or HOAc), nitric acid (HNO3), and benzoic acid (C6H5CO2H) are all monoprotic acids.
Several important acids can be classified as polyprotic acids, which can lose more than one H+ ion when they act as Brnsted acids. Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4) have two acidic hydrogen atoms. Triprotic acids, such as phosphoric acid (H3PO4) and citric acid (C6H8O7), have three.
There is usually a large difference in the ease with which these acids lose the first and second (or second and third) protons. When sulfuric acid is classified as a strong acid, students often assume that it loses both of its protons when it reacts with water. That isn't a legitimate assumption. Sulfuric acid is a strong acid because Ka for the loss of the first proton is much larger than 1. We therefore assume that essentially all the H2SO4 molecules in an aqueous solution lose the first proton to form the HSO4-, or hydrogen sulfate, ion.
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)
Ka1 = 1 x 103
But Ka for the loss of the second proton is only 10-2 and only 10% of the H2SO4 molecules in a 1 Msolution lose a second proton.
HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq)
Ka2 = 1.2 x 10-2
H2SO4 only loses both H+ ions when it reacts with a base, such as ammonia.
The table below gives values of Ka for some common polyprotic acids. The large difference between the values of Ka for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a timean assumption known as stepwise dissociation.
Acid-Dissociation Equilibrium Constants for Common Polyprotic Acids
Acid
Ka1
Ka2
Ka3
sulfuric acid (H2SO4)
1.0 x 103
1.2 x 10-2
chromic acid (H2CrO4)
9.6
3.2 x 10-7
oxalic acid (H2C2O4)
5.4 x 10-2
5.4 x 10-5
sulfurous acid (H2SO3)
1.7 x 10-2
6.4 x 10-8
phosphoric acid (H3PO4)
7.1 x 10-3
6.3 x 10-8
4.2 x 10-13
glycine (C2H6NO2)
4.5 x 10-3
2.5 x 10-10
citric acid (C6H8O7)
7.5 x 10-4
1.7 x 10-5
4.0 x 10-7
carbonic acid (H2CO3)
4.5 x 10-7
4.7 x 10-11
hydrogen sulfide (H2S)
1.0 x 10-7
1.3 x 10-13
Let's look at the consequence of the assumption that polyprotic acids lose protons one step at a time by examining the chemistry of a saturated solution of H2S in water.
Hydrogen sulfide is the foul-smelling gas that gives rotten eggs their unpleasant odor. It is an excellent source of the S2- ion, however, and is therefore commonly used in introductory chemistry laboratories. H2S is a weak acid that dissociates in steps. Some of the H2S molecules lose a proton in the first step to form the HS-, or hydrogen sulfide, ion.
First step:
H2S(aq) + H2O(l) H3O+(aq) + HS-(aq)
A small fraction of the HS- ions formed in this reaction then go on to lose another H+ ion in a second step.
Second step:
HS-(aq) + H2O(l) H3O+(aq) + S2-(aq)
Since there are two steps in this reaction, we can write two equilibrium constant expressions.
Although each of these equations contains three terms, there are only four unknowns[H3O+], [H2S], [HS-], and [S2-] because the [H3O+] and [HS-] terms appear in both equations. The [H3O+] term represents the total H3O+ ion concentration from both steps and therefore must have the same value in both equations. Similarly, the [HS-] term, which represents the balance between the HS- ions formed in the first step and the HS- ions consumed in the second step, must have the same value for both equations.
Four equations are needed to solve for four unknowns. We already have two equations: the Ka1 andKa2 expressions. We are going to have to find either two more equations or a pair of assumptions that can generate two equations. We can base one assumption on the fact that the value of Ka1 for this acid is almost a million times larger than the value of Ka2.
Ka1 >> Ka2
This means that only a small fraction of the HS- ions formed in the first step go on to dissociate in the second step. If this is true, most of the H3O+ ions in this solution come from the dissociation of H2S, and most of the HS- ions formed in this reaction PSS remain in solution. As a result, we can assume that the H3O+ and HS- ion concentrations are more or less equal.
First assumption:
[H3O+][HS-]
We need one more equation, and therefore one more assumption. Note that H2S is a weak acid (Ka1= 1.0 x 10-7, Ka2 = 1.3 x 10-13). Thus, we can assume that most of the H2S that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H2S is approximately equal to the initial concentration.
Second assumption:
[H2S] CH2S
We now have four equations in four unknowns.
[H3O+] [HS-]
[H2S] CH2S
Since there is always a unique solution to four equations in four unknowns, we are now ready to calculate the H3O+, H2S, HS-, and S2- concentrations at equilibrium in a saturated solution of H2S in water. All we need to know is that a saturated solution of H2S in water has an initial concentration of about 0.10 M.
Because Ka1 is so much larger than Ka2 for this acid, we can work with the equilibrium expression for the first step without worrying about the second step for the moment. We therefore start with the expression for Ka1 for this acid.
We then invoke one of our assumptions.
[H2S] CH2S 0.10 M
Substituting this approximation into the Ka1 expression gives the following equation.
We then invoke the other assumption.
[H3O+] [HS-]C
Substituting this approximation into the Ka1 expression gives the following result.
We now solve this approximate equation for C.
C 1.0 x 10-4
If our two assumptions are valid, we are three-fourths of the way to our goal. We know the H2S, H3O+, and HS- concentrations.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
Having extracted the values of three unknowns from the first equilibrium expression, we turn to the second equilibrium expression.
Substituting the known values of the H3O+ and HS- ion concentrations into this expression gives the following equation.
Because the equilibrium concentrations of the H3O+ and HS- ions are more or less the same, the S2-ion concentration at equilibrium is approximately equal to the value of Ka2 for this acid.
[S2-] 1.3 x 10-13 M
It is now time to check our assumptions. Is the dissociation of H2S small compared with the initial concentration? Yes. The HS- and H3O+ ion concentrations obtained from this calculation are 1.0 x 10-4 M, which is 0.1% of the initial concentration of H2S. The following assumption is therefore valid.
[H2S] CH2S 0.10 M
Is the difference between the S2- and HS- ion concentrations large enough to allow us to assume that essentially all of the H3O+ ions at equilibrium are formed in the first step and that essentially all of the HS- ions formed in this step remain in solution? Yes. The S2- ion concentration obtained from this calculation is 109 times smaller than the HS- ion concentration. Thus, our other assumption is also valid.
[H3O+] [HS-]
We can therefore summarize the concentrations of the various components of this equilibrium as follows.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
[S2-] 1.3 x 10-13 M
Diprotic Bases
By: Sandra May Dela Cruz
A diprotic acid is an acid such as H2SO4 (sulfuric acid) that contains within its molecular structure two hydrogen atoms per molecule capable of dissociating (i.e. ionizable) in water. The complete dissociation of diprotic acids is of the same form as sulfuric acid:
H2SO4 → H+(aq) + HSO4−(aq) Ka = 1 × 103
HSO4− → H+(aq) + SO42−(aq) Ka = 1 × 10−2
The dissociation does not happen all at once due to the two stages of dissociation having different Ka values. The first dissociation will, in the case of sulfuric acid, occur completely, but the second one will not. Diprotic acids are of particular note in regards to titrationexperiments, where a pH versus titrant volume curve will clearly show two equivalence points for the acid. This occurs because the two ionization capable hydrogen atoms on the acid molecule do not leave the acid at the same time.
Common diprotic acids include malic acid, found in apples and cherries, and tartaric acid, found in grapes and pineapples. Another diprotic acid is uric acid, which has a pKa1 of 5.4 and a pKa2 of 10.3, and is therefore singly charged at physiological pH.
The techniques we have used with diprotic acids can be extended to diprotic bases. The only challenge is calculating the values of Kb for the base.
Example: Let's calculate the H2CO3, HCO3-, CO32-, and OH- concentrations at equilibrium in a solution that is initially 0.10 M in Na2CO3. (H2CO3: Ka1 = 4.5 x 10-7; Ka2 = 4.7 x 10-11)
Because it is a salt, sodium carbonate dissociates into its ions when it dissolves in water.
H2O Na2CO3(aq)---2 Na+CO32-(aq)
The carbonate ion then acts as a base toward water, picking up a pair of protons (one at a time) to form the bicarbonate ion, HCO3- ion, and then eventually carbonic acid, H2CO3.
CO3 2-(aq) + H2O(l) HCO3-(aq) + OH-(aq)
Kb1 = ?
HCO3-(aq) + H2O(l) H2CO3(aq) + OH-(aq)
Kb2 = ?
The first step in solving this problem involves determining the values of Kb1 and Kb2 for the carbonate ion. We start by comparing the Kb expressions for the carbonate ion with the Kaexpressions for carbonic acid.
The expressions for Kb1 and Ka2 have something in commonthey both depend on the concentrations of the HCO3- and CO32- ions. The expressions for Kb2 and Ka1 also have something in commonthey both depend on the HCO3- and H2CO3 concentrations. We can therefore calculateKb1 from Ka2 and Kb2 from Ka1.
We start by multiplying the top and bottom of the Ka1 expression by the OH- ion concentration to introduce the [OH-] term.
We then group terms in this equation as follows.
The first term in this equation is the inverse of the Kb2 expression, and the second term is the Kwexpression.
Rearranging this equation gives the following result.
Ka1Kb2 = Kw
Similarly, we can multiply the top and bottom of the Ka2 expression by the OH-ion concentration.
Collecting terms gives the following equation.
The first term in this equation is the inverse of Kb1, and the second term is Kw.
This equation can therefore be rearranged as follows.
Ka2Kb1 = Kw
We can now calculate the values of Kb1 and Kb2 for the carbonate ion.
We are finally ready to do the calculations. We start with the Kb1 expression because the CO32- ion is the strongest base in this solution and therefore the best source of the OH- ion.
The difference between Kb1 and Kb2 for the carbonate ion is large enough to suggest that most of the OH- ions come from this step and most of the HCO3- ions formed in this reaction remain in solution.
[OH-] [HCO3-] C
The value of Kb1 is small enough to assume that C is small compared with the initial concentration of the carbonate ion. If this is true, the concentration of the CO32- ion at equilibrium will be roughly equal to the initial concentration of Na2CO3.
[CO32-] CNa2CO3
Substituting this information into the Kb1 expression gives the following result.
This approximate equation can now be solved for C.
C 0.0046 M
We then use this value of C to calculate the equilibrium concentrations of the OH-, HCO3-, and CO32- ions.
[CO32-] = 0.10 - C 0.095 M
[OH-] [HCO3-] C 0.0046 M
We now turn to the Kb2 expression.
Substituting what we know about the OH- and HCO3- ion concentrations into this equation gives the following result.
According to this equation, the H2CO3 concentration at equilibrium is approximately equal to Kb2 for the carbonate ion.
[H2CO3] 2.2 x 10-8 M
Summarizing the results of our calculations allows us to test the assumptions made generating these results.
[CO32-] 0.095 M
[OH-] [HCO3-] 4.6 x 10-3 M
[H2CO3] 2.2 x 10-8 M
All of our assumptions are valid. The extent of the reaction between the CO32- ion and water to give the HCO3- ion is less than 5% of the initial concentration of Na2CO3. Furthermore, most of the OH-ion comes from the first step, and most of the HCO3- ion formed in this step remains in solution.
By: Sandra May Dela Cruz
A diprotic acid is an acid such as H2SO4 (sulfuric acid) that contains within its molecular structure two hydrogen atoms per molecule capable of dissociating (i.e. ionizable) in water. The complete dissociation of diprotic acids is of the same form as sulfuric acid:
H2SO4 → H+(aq) + HSO4−(aq) Ka = 1 × 103
HSO4− → H+(aq) + SO42−(aq) Ka = 1 × 10−2
The dissociation does not happen all at once due to the two stages of dissociation having different Ka values. The first dissociation will, in the case of sulfuric acid, occur completely, but the second one will not. Diprotic acids are of particular note in regards to titrationexperiments, where a pH versus titrant volume curve will clearly show two equivalence points for the acid. This occurs because the two ionization capable hydrogen atoms on the acid molecule do not leave the acid at the same time.
Common diprotic acids include malic acid, found in apples and cherries, and tartaric acid, found in grapes and pineapples. Another diprotic acid is uric acid, which has a pKa1 of 5.4 and a pKa2 of 10.3, and is therefore singly charged at physiological pH.
The techniques we have used with diprotic acids can be extended to diprotic bases. The only challenge is calculating the values of Kb for the base.
Example: Let's calculate the H2CO3, HCO3-, CO32-, and OH- concentrations at equilibrium in a solution that is initially 0.10 M in Na2CO3. (H2CO3: Ka1 = 4.5 x 10-7; Ka2 = 4.7 x 10-11)
Because it is a salt, sodium carbonate dissociates into its ions when it dissolves in water.
H2O Na2CO3(aq)---2 Na+CO32-(aq)
The carbonate ion then acts as a base toward water, picking up a pair of protons (one at a time) to form the bicarbonate ion, HCO3- ion, and then eventually carbonic acid, H2CO3.
CO3 2-(aq) + H2O(l) HCO3-(aq) + OH-(aq)
Kb1 = ?
HCO3-(aq) + H2O(l) H2CO3(aq) + OH-(aq)
Kb2 = ?
The first step in solving this problem involves determining the values of Kb1 and Kb2 for the carbonate ion. We start by comparing the Kb expressions for the carbonate ion with the Kaexpressions for carbonic acid.
The expressions for Kb1 and Ka2 have something in commonthey both depend on the concentrations of the HCO3- and CO32- ions. The expressions for Kb2 and Ka1 also have something in commonthey both depend on the HCO3- and H2CO3 concentrations. We can therefore calculateKb1 from Ka2 and Kb2 from Ka1.
We start by multiplying the top and bottom of the Ka1 expression by the OH- ion concentration to introduce the [OH-] term.
We then group terms in this equation as follows.
The first term in this equation is the inverse of the Kb2 expression, and the second term is the Kwexpression.
Rearranging this equation gives the following result.
Ka1Kb2 = Kw
Similarly, we can multiply the top and bottom of the Ka2 expression by the OH-ion concentration.
Collecting terms gives the following equation.
The first term in this equation is the inverse of Kb1, and the second term is Kw.
This equation can therefore be rearranged as follows.
Ka2Kb1 = Kw
We can now calculate the values of Kb1 and Kb2 for the carbonate ion.
We are finally ready to do the calculations. We start with the Kb1 expression because the CO32- ion is the strongest base in this solution and therefore the best source of the OH- ion.
The difference between Kb1 and Kb2 for the carbonate ion is large enough to suggest that most of the OH- ions come from this step and most of the HCO3- ions formed in this reaction remain in solution.
[OH-] [HCO3-] C
The value of Kb1 is small enough to assume that C is small compared with the initial concentration of the carbonate ion. If this is true, the concentration of the CO32- ion at equilibrium will be roughly equal to the initial concentration of Na2CO3.
[CO32-] CNa2CO3
Substituting this information into the Kb1 expression gives the following result.
This approximate equation can now be solved for C.
C 0.0046 M
We then use this value of C to calculate the equilibrium concentrations of the OH-, HCO3-, and CO32- ions.
[CO32-] = 0.10 - C 0.095 M
[OH-] [HCO3-] C 0.0046 M
We now turn to the Kb2 expression.
Substituting what we know about the OH- and HCO3- ion concentrations into this equation gives the following result.
According to this equation, the H2CO3 concentration at equilibrium is approximately equal to Kb2 for the carbonate ion.
[H2CO3] 2.2 x 10-8 M
Summarizing the results of our calculations allows us to test the assumptions made generating these results.
[CO32-] 0.095 M
[OH-] [HCO3-] 4.6 x 10-3 M
[H2CO3] 2.2 x 10-8 M
All of our assumptions are valid. The extent of the reaction between the CO32- ion and water to give the HCO3- ion is less than 5% of the initial concentration of Na2CO3. Furthermore, most of the OH-ion comes from the first step, and most of the HCO3- ion formed in this step remains in solution.
Diprotic Acids
By: Daryl Dave Crisostomo
A diprotic acid is an acid that can donate two proton or hydrogen atom per molecule to an aqueous solution.
The acid equilibrium problems discussed so far have focused on a family of compounds known as monoprotic acids. Each of these acids has a single H+ ion, or proton, it can donate when it acts as a Brnsted acid. Hydrochloric acid (HCl), acetic acid (CH3CO2H or HOAc), nitric acid (HNO3), and benzoic acid (C6H5CO2H) are all monoprotic acids.
Several important acids can be classified as polyprotic acids, which can lose more than one H+ ion when they act as Brnsted acids. Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4) have two acidic hydrogen atoms. Triprotic acids, such as phosphoric acid (H3PO4) and citric acid (C6H8O7), have three.
There is usually a large difference in the ease with which these acids lose the first and second (or second and third) protons. When sulfuric acid is classified as a strong acid, students often assume that it loses both of its protons when it reacts with water. That isn't a legitimate assumption. Sulfuric acid is a strong acid because Ka for the loss of the first proton is much larger than 1. We therefore assume that essentially all the H2SO4 molecules in an aqueous solution lose the first proton to form the HSO4-, or hydrogen sulfate, ion.
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)
Ka1 = 1 x 103
But Ka for the loss of the second proton is only 10-2 and only 10% of the H2SO4 molecules in a 1 Msolution lose a second proton.
HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq)
Ka2 = 1.2 x 10-2
H2SO4 only loses both H+ ions when it reacts with a base, such as ammonia.
The table below gives values of Ka for some common polyprotic acids. The large difference between the values of Ka for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a timean assumption known as stepwise dissociation.
Acid-Dissociation Equilibrium Constants for Common Polyprotic Acids
Acid
Ka1
Ka2
Ka3
sulfuric acid (H2SO4)
1.0 x 103
1.2 x 10-2
chromic acid (H2CrO4)
9.6
3.2 x 10-7
oxalic acid (H2C2O4)
5.4 x 10-2
5.4 x 10-5
sulfurous acid (H2SO3)
1.7 x 10-2
6.4 x 10-8
phosphoric acid (H3PO4)
7.1 x 10-3
6.3 x 10-8
4.2 x 10-13
glycine (C2H6NO2)
4.5 x 10-3
2.5 x 10-10
citric acid (C6H8O7)
7.5 x 10-4
1.7 x 10-5
4.0 x 10-7
carbonic acid (H2CO3)
4.5 x 10-7
4.7 x 10-11
hydrogen sulfide (H2S)
1.0 x 10-7
1.3 x 10-13
Let's look at the consequence of the assumption that polyprotic acids lose protons one step at a time by examining the chemistry of a saturated solution of H2S in water.
Hydrogen sulfide is the foul-smelling gas that gives rotten eggs their unpleasant odor. It is an excellent source of the S2- ion, however, and is therefore commonly used in introductory chemistry laboratories. H2S is a weak acid that dissociates in steps. Some of the H2S molecules lose a proton in the first step to form the HS-, or hydrogen sulfide, ion.
First step:
H2S(aq) + H2O(l) H3O+(aq) + HS-(aq)
A small fraction of the HS- ions formed in this reaction then go on to lose another H+ ion in a second step.
Second step:
HS-(aq) + H2O(l) H3O+(aq) + S2-(aq)
Since there are two steps in this reaction, we can write two equilibrium constant expressions.
Although each of these equations contains three terms, there are only four unknowns[H3O+], [H2S], [HS-], and [S2-] because the [H3O+] and [HS-] terms appear in both equations. The [H3O+] term represents the total H3O+ ion concentration from both steps and therefore must have the same value in both equations. Similarly, the [HS-] term, which represents the balance between the HS- ions formed in the first step and the HS- ions consumed in the second step, must have the same value for both equations.
Four equations are needed to solve for four unknowns. We already have two equations: the Ka1 andKa2 expressions. We are going to have to find either two more equations or a pair of assumptions that can generate two equations. We can base one assumption on the fact that the value of Ka1 for this acid is almost a million times larger than the value of Ka2.
Ka1 >> Ka2
This means that only a small fraction of the HS- ions formed in the first step go on to dissociate in the second step. If this is true, most of the H3O+ ions in this solution come from the dissociation of H2S, and most of the HS- ions formed in this reaction PSS remain in solution. As a result, we can assume that the H3O+ and HS- ion concentrations are more or less equal.
First assumption:
[H3O+][HS-]
We need one more equation, and therefore one more assumption. Note that H2S is a weak acid (Ka1= 1.0 x 10-7, Ka2 = 1.3 x 10-13). Thus, we can assume that most of the H2S that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H2S is approximately equal to the initial concentration.
Second assumption:
[H2S] CH2S
We now have four equations in four unknowns.
[H3O+] [HS-]
[H2S] CH2S
Since there is always a unique solution to four equations in four unknowns, we are now ready to calculate the H3O+, H2S, HS-, and S2- concentrations at equilibrium in a saturated solution of H2S in water. All we need to know is that a saturated solution of H2S in water has an initial concentration of about 0.10 M.
Because Ka1 is so much larger than Ka2 for this acid, we can work with the equilibrium expression for the first step without worrying about the second step for the moment. We therefore start with the expression for Ka1 for this acid.
We then invoke one of our assumptions.
[H2S] CH2S 0.10 M
Substituting this approximation into the Ka1 expression gives the following equation.
We then invoke the other assumption.
[H3O+] [HS-]C
Substituting this approximation into the Ka1 expression gives the following result.
We now solve this approximate equation for C.
C 1.0 x 10-4
If our two assumptions are valid, we are three-fourths of the way to our goal. We know the H2S, H3O+, and HS- concentrations.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
Having extracted the values of three unknowns from the first equilibrium expression, we turn to the second equilibrium expression.
Substituting the known values of the H3O+ and HS- ion concentrations into this expression gives the following equation.
Because the equilibrium concentrations of the H3O+ and HS- ions are more or less the same, the S2-ion concentration at equilibrium is approximately equal to the value of Ka2 for this acid.
[S2-] 1.3 x 10-13 M
It is now time to check our assumptions. Is the dissociation of H2S small compared with the initial concentration? Yes. The HS- and H3O+ ion concentrations obtained from this calculation are 1.0 x 10-4 M, which is 0.1% of the initial concentration of H2S. The following assumption is therefore valid.
[H2S] CH2S 0.10 M
Is the difference between the S2- and HS- ion concentrations large enough to allow us to assume that essentially all of the H3O+ ions at equilibrium are formed in the first step and that essentially all of the HS- ions formed in this step remain in solution? Yes. The S2- ion concentration obtained from this calculation is 109 times smaller than the HS- ion concentration. Thus, our other assumption is also valid.
[H3O+] [HS-]
We can therefore summarize the concentrations of the various components of this equilibrium as follows.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
[S2-] 1.3 x 10-13 M
By: Daryl Dave Crisostomo
A diprotic acid is an acid that can donate two proton or hydrogen atom per molecule to an aqueous solution.
The acid equilibrium problems discussed so far have focused on a family of compounds known as monoprotic acids. Each of these acids has a single H+ ion, or proton, it can donate when it acts as a Brnsted acid. Hydrochloric acid (HCl), acetic acid (CH3CO2H or HOAc), nitric acid (HNO3), and benzoic acid (C6H5CO2H) are all monoprotic acids.
Several important acids can be classified as polyprotic acids, which can lose more than one H+ ion when they act as Brnsted acids. Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4) have two acidic hydrogen atoms. Triprotic acids, such as phosphoric acid (H3PO4) and citric acid (C6H8O7), have three.
There is usually a large difference in the ease with which these acids lose the first and second (or second and third) protons. When sulfuric acid is classified as a strong acid, students often assume that it loses both of its protons when it reacts with water. That isn't a legitimate assumption. Sulfuric acid is a strong acid because Ka for the loss of the first proton is much larger than 1. We therefore assume that essentially all the H2SO4 molecules in an aqueous solution lose the first proton to form the HSO4-, or hydrogen sulfate, ion.
H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq)
Ka1 = 1 x 103
But Ka for the loss of the second proton is only 10-2 and only 10% of the H2SO4 molecules in a 1 Msolution lose a second proton.
HSO4-(aq) + H2O(l) H3O+(aq) + SO42-(aq)
Ka2 = 1.2 x 10-2
H2SO4 only loses both H+ ions when it reacts with a base, such as ammonia.
The table below gives values of Ka for some common polyprotic acids. The large difference between the values of Ka for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a timean assumption known as stepwise dissociation.
Acid-Dissociation Equilibrium Constants for Common Polyprotic Acids
Acid
Ka1
Ka2
Ka3
sulfuric acid (H2SO4)
1.0 x 103
1.2 x 10-2
chromic acid (H2CrO4)
9.6
3.2 x 10-7
oxalic acid (H2C2O4)
5.4 x 10-2
5.4 x 10-5
sulfurous acid (H2SO3)
1.7 x 10-2
6.4 x 10-8
phosphoric acid (H3PO4)
7.1 x 10-3
6.3 x 10-8
4.2 x 10-13
glycine (C2H6NO2)
4.5 x 10-3
2.5 x 10-10
citric acid (C6H8O7)
7.5 x 10-4
1.7 x 10-5
4.0 x 10-7
carbonic acid (H2CO3)
4.5 x 10-7
4.7 x 10-11
hydrogen sulfide (H2S)
1.0 x 10-7
1.3 x 10-13
Let's look at the consequence of the assumption that polyprotic acids lose protons one step at a time by examining the chemistry of a saturated solution of H2S in water.
Hydrogen sulfide is the foul-smelling gas that gives rotten eggs their unpleasant odor. It is an excellent source of the S2- ion, however, and is therefore commonly used in introductory chemistry laboratories. H2S is a weak acid that dissociates in steps. Some of the H2S molecules lose a proton in the first step to form the HS-, or hydrogen sulfide, ion.
First step:
H2S(aq) + H2O(l) H3O+(aq) + HS-(aq)
A small fraction of the HS- ions formed in this reaction then go on to lose another H+ ion in a second step.
Second step:
HS-(aq) + H2O(l) H3O+(aq) + S2-(aq)
Since there are two steps in this reaction, we can write two equilibrium constant expressions.
Although each of these equations contains three terms, there are only four unknowns[H3O+], [H2S], [HS-], and [S2-] because the [H3O+] and [HS-] terms appear in both equations. The [H3O+] term represents the total H3O+ ion concentration from both steps and therefore must have the same value in both equations. Similarly, the [HS-] term, which represents the balance between the HS- ions formed in the first step and the HS- ions consumed in the second step, must have the same value for both equations.
Four equations are needed to solve for four unknowns. We already have two equations: the Ka1 andKa2 expressions. We are going to have to find either two more equations or a pair of assumptions that can generate two equations. We can base one assumption on the fact that the value of Ka1 for this acid is almost a million times larger than the value of Ka2.
Ka1 >> Ka2
This means that only a small fraction of the HS- ions formed in the first step go on to dissociate in the second step. If this is true, most of the H3O+ ions in this solution come from the dissociation of H2S, and most of the HS- ions formed in this reaction PSS remain in solution. As a result, we can assume that the H3O+ and HS- ion concentrations are more or less equal.
First assumption:
[H3O+][HS-]
We need one more equation, and therefore one more assumption. Note that H2S is a weak acid (Ka1= 1.0 x 10-7, Ka2 = 1.3 x 10-13). Thus, we can assume that most of the H2S that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H2S is approximately equal to the initial concentration.
Second assumption:
[H2S] CH2S
We now have four equations in four unknowns.
[H3O+] [HS-]
[H2S] CH2S
Since there is always a unique solution to four equations in four unknowns, we are now ready to calculate the H3O+, H2S, HS-, and S2- concentrations at equilibrium in a saturated solution of H2S in water. All we need to know is that a saturated solution of H2S in water has an initial concentration of about 0.10 M.
Because Ka1 is so much larger than Ka2 for this acid, we can work with the equilibrium expression for the first step without worrying about the second step for the moment. We therefore start with the expression for Ka1 for this acid.
We then invoke one of our assumptions.
[H2S] CH2S 0.10 M
Substituting this approximation into the Ka1 expression gives the following equation.
We then invoke the other assumption.
[H3O+] [HS-]C
Substituting this approximation into the Ka1 expression gives the following result.
We now solve this approximate equation for C.
C 1.0 x 10-4
If our two assumptions are valid, we are three-fourths of the way to our goal. We know the H2S, H3O+, and HS- concentrations.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
Having extracted the values of three unknowns from the first equilibrium expression, we turn to the second equilibrium expression.
Substituting the known values of the H3O+ and HS- ion concentrations into this expression gives the following equation.
Because the equilibrium concentrations of the H3O+ and HS- ions are more or less the same, the S2-ion concentration at equilibrium is approximately equal to the value of Ka2 for this acid.
[S2-] 1.3 x 10-13 M
It is now time to check our assumptions. Is the dissociation of H2S small compared with the initial concentration? Yes. The HS- and H3O+ ion concentrations obtained from this calculation are 1.0 x 10-4 M, which is 0.1% of the initial concentration of H2S. The following assumption is therefore valid.
[H2S] CH2S 0.10 M
Is the difference between the S2- and HS- ion concentrations large enough to allow us to assume that essentially all of the H3O+ ions at equilibrium are formed in the first step and that essentially all of the HS- ions formed in this step remain in solution? Yes. The S2- ion concentration obtained from this calculation is 109 times smaller than the HS- ion concentration. Thus, our other assumption is also valid.
[H3O+] [HS-]
We can therefore summarize the concentrations of the various components of this equilibrium as follows.
[H2S] 0.10 M
[H3O+] [HS-] 1.0 x 10-4 M
[S2-] 1.3 x 10-13 M