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pH of Polyprotic Weak Acids Tutorial

Key Concepts

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Defining Polyprotic Weak Acids

A monoprotic acid is a Brønsted-Lowry acid that is able to dontate only one proton 2

Example: HA H+ + A-

HA is a monoprotic acid because it only has one proton to donate.

HA is a weak acid because it does not completely dissociate (ionise), that is, the molecular form of the acid is in equilibrium with the ions produced by its partial dissociation.

A polyprotic acid is a Brønsted-Lowry acid that is able to dontate more than one proton3

A monoprotic acid is NOT an example of a polyprotic acid.

A diprotic acid is an example of a polyprotic acid which is a Brønsted-Lowry acid able to donate two protons4

A diprotic acid dissociates (or ionises) in stages, donating one proton in each stage

Stage 1: H2A H+ + HA-

Stage 2: HA- H+ + A2-

H2A is a diprotic acid because it can donate 2 protons.

H2A is a weak acid because it does not completely dissociate (ionise), that is, the molecular form of the acid is in equilibrium with the ions produced by its partial dissociation.
Each of the following solute species will be present in the solution at equilibrium:

  • H2A (undissociated, molecular species)
  • HA-
  • A2-
  • H+ (protons, also known as hydrons)

A triprotic acid is is an example of a polyprotic acid which is a Brønsted-Lowry acid able to donate three protons5

A triprotic acid dissociates (or ionises) in stages, donating one proton in each stage

Stage 1: H3A H+ + H2A-

Stage 2: H2A- H+ + HA2-

Stage 2: HA2- H+ + A3-

H3A is a triprotic acid because it can donate 3 protons.

H3A is a weak acid because it does not completely dissociate (ionise), that is, the molecular form of the acid is in equilibrium with the ions produced by its partial dissociation.
Each of the following solute species will be present in the solution at equilibrium:

  • H3A (undissociated, molecular species)
  • H2A-
  • HA2-
  • A3-
  • H+ (protons, also known as hydrons)

Examples of polyprotic weak acids in aqueous solution are:

NameMolecular
Formula
TypeDissociation Stages
sulfurous acidH2SO3(aq)a diprotic acidI. H2SO3(aq) H+(aq) + HSO3-(aq)
II. HSO3-(aq) H+(aq) + SO32-(aq)

carbonic acidH2CO3(aq)a diprotic acidI. H2CO3(aq) H+(aq) + HCO3-(aq)
II. HCO3-(aq) H+(aq) + CO32-(aq)

hydrosulfuric acid
(hydrogen sulfide)
H2S(aq)a diprotic acidI. H2S(aq) H+(aq) + HS-(aq)
II. HS-(aq) H+(aq) + S2-(aq)

oxalic acidH2C2O4(aq)a diprotic acidI. H2C2O4(aq) H+(aq) + HC2O4-(aq)
II. HC2O4-(aq) H+(aq) + C2O42-(aq)

phosphoric acidH3PO4(aq)a triprotic acidI. H3PO4(aq) H+(aq) + H2PO4-(aq)
II. H2PO4-(aq) H+(aq) + HPO42-(aq)
III. HPO42-(aq) H+(aq) + PO43-(aq)

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Ka and Polyprotic Weak Acids

There is a corresponding acid dissociation (ionisation) constant, Ka, for each stage.
For the dissociation of acid HxA:

acid proton + conjugate base
HxA H+ + Hx-1A-
Ka = [protons][conjugate base]
[acid]

A subscript number adjacent to the "a" in the symbol Ka refers to the stage:

Examples of the acid dissociation expression for some polyprotic weak acids in aqueous solution are:

NameMolecular
Formula
TypeDissociation StagesAcid Dissociation Expression
sulfurous acidH2SO3(aq)a diprotic acidI. H2SO3(aq) H+(aq) + HSO3-(aq)

 

II. HSO3-(aq) H+(aq) + SO32-(aq)

Ka1 = [H+][HSO3-]
[H2SO3]

Ka2 = [H+][SO32-]
[HSO3-]

carbonic acidH2CO3(aq)a diprotic acidI. H2CO3(aq) H+(aq) + HCO3-(aq)

 

II. HCO3-(aq) H+(aq) + CO32-(aq)

Ka1 = [H+][HCO3-]
[H2CO3]

Ka2 = [H+][CO32-]
[HCO3-]

hydrosulfuric acid
(hydrogen sulfide)
H2S(aq)a diprotic acidI. H2S(aq) H+(aq) + HS-(aq)

 

II. HS-(aq) H+(aq) + S2-(aq)

Ka1 = [H+][HS-]
[H2S]

Ka2 = [H+][S2-]
[HS-]

oxalic acidH2C2O4(aq)a diprotic acidI. H2C2O4(aq) H+(aq) + HC2O4-(aq)

 

II. HC2O4-(aq) H+(aq) + C2O42-(aq)

Ka1 = [H+][HC2O4-]
[H2C2O4]

Ka2 = [H+][C2O42-]
[HC2O4-]

phosphoric acidH3PO4(aq)a triprotic acidI. H3PO4(aq) H+(aq) + H2PO4-(aq)

 

II. H2PO4-(aq) H+(aq) + HPO42-(aq)

 

III. HPO42-(aq) H+(aq) + PO43-(aq)

Ka1 = [H+][H2PO4-]
[H3PO4]

Ka2 = [H+][HPO42-]
[H2PO4-]

Ka3 = [H+][PO43-]
[HPO42-]

Consider the addition of 1 mole of the weak diprotic acid H2A to 1 L of water.
First, some of the molecular H2A, say 1%, will dissociate, that is, 99% of the H2A is undissociated.
We can use a R.I.C.E. Table to determine the equilibrium concentrations of each each species in solution:

R.I.C.E. Table
Reaction: H2A(aq) H+(aq) + HA-(aq)
Initial Concentration:
mol L-1
1
(given)
  0   0
Change in Concentration:
mol L-1
-1% of 1 = 0.01   +0.01   +0.01
Equilibrium Concentration:
mol L-1
1 − 0.01 = 0.99   0 +0.01 = 0.01   0 +0.01 = 0.01

We can calculate the first acid dissociation constant, Ka1, for this first stage:

Ka1 = [H+][HA-]
[H2A]
= [0.01][0.01]
[0.99]
= 1 × 10-4

Next, some of the HA-(aq) produced above will dissociate, but not much!
We stated that H2A was a weak acid, that is, it doesn't dissociate very much, but this also means that for the reaction:

H2A(aq) H+(aq) + HA-(aq)

the equiliubrium position lies very far to the left, so there is even less tendency for HA-(aq) to dissociate.
Therefore for the reaction:
HA-(aq) H+(aq) + A2-(aq)

the equilibrium position will lie even further to the left.
Let's say only 0.1% of the HA-(aq) dissociates (ionises).
Once again we can use a R.I.C.E. Table to determine the equilibrium concentrations of these species:
R.I.C.E. Table
Reaction: HA-(aq) H+(aq) + A2-(aq)
Initial Concentration:
mol L-1
0.01
(calculated above)
  0.01
(calculated above)
  0
Change in Concentration:
mol L-1
-0.1% of 0.01 = 1 × 10-5   +1 × 10-5   +1 × 10-5
Equilibrium Concentration:
mol L-1
0.01 − 1 × 10-5
= 9.99 × 10-3
  0.01 + 1 × 10-5
= 0.01001
  0 + 1 × 10-5
= 1 × 10-5

We can calculate the second acid dissociation constant, Ka2, for this first stage:

Ka2 = [H+][A2-]
[HA-]
= [0.01001][1 × 10-5]
[9.99 × 10-3]
= 1 × 10-5

Note: Ka1 (10-4) > Ka2 (10-5) therefore H2A is a stronger acid (more dissociated) that HA-.

For polyprotic acids, the first dissociation (ionisation) constant, Ka1, is always greater than the second dissociation (ionisation) constant, Ka2, which is greater than the third dissociation (ionisation constant), Ka3, etc:

Examples of the acid dissociation constants at 25o for some polyprotic weak acids in aqueous solution are:

NameMolecular
Formula
TypeDissociation StagesKa
25oC
sulfurous acidH2SO3diprotic
acid
I. H2SO3(aq) H+(aq) + HSO3-(aq)

 

II. HSO3-(aq) H+(aq) + SO32-(aq)

Ka1 = [H+][HSO3-]
[H2SO3]
= 1.3 × 10-2

Ka2 = [H+][SO32-]
[HSO3-]
= 6.3 × 10-8

carbonic acidH2CO3(aq)diprotic
acid
I. H2CO3(aq) H+(aq) + HCO3-(aq)

 

II. HCO3-(aq) H+(aq) + CO32-(aq)

Ka1 = [H+][HCO3-]
[H2CO3]
= 4.2 × 10-7

Ka2 = [H+][CO32-]
[HCO3-]
= 5.6 × 10-11

hydrosulfuric acid
(hydrogen sulfide)
H2S(aq)diprotic
acid
I. H2S(aq) H+(aq) + HS-(aq)

 

II. HS-(aq) H+(aq) + S2-(aq)

Ka1 = [H+][HS-]
[H2S]
= 1.1 × 10-7

Ka2 = [H+][S2-]
[HS-]
= 1.0 × 10-14

oxalic acidH2C2O4(aq)diprotic
acid
I. H2C2O4(aq) H+(aq) + HC2O4-(aq)

 

II. HC2O4-(aq) H+(aq) + C2O42-(aq)

Ka1 = [H+][HC2O4-]
[H2C2O4]
= 5.4 × 10-2

Ka2 = [H+][C2O42-]
[HC2O4-]
= 5.0 × 10-5

phosphoric acidH3PO4(aq)triprotic
acid
I. H3PO4(aq) H+(aq) + H2PO4-(aq)

 

II. H2PO4-(aq) H+(aq) + HPO42-(aq)

 

III. HPO42-(aq) H+(aq) + PO43-(aq)

Ka1 = [H+][H2PO4-]
[H3PO4]
= 7.6 × 10-3

Ka2 = [H+][HPO42-]
[H2PO4-]
= 6.3 × 10-8

Ka3 = [H+][PO43-]
[HPO42-]
= 4.4 × 10-13

diprotic acid largest Ka smallest Ka
general: H2A: Ka1 > Ka2
H2CO3(aq) :
(25oC)
4.1 × 10-7 > 5.6 × 10-11
 
triprotic acid largest Ka → → → → smallest Ka
general: H3A:Ka1 > Ka2 > Ka3
H3PO4(aq) :
(25oC)
7.6 × 10-3 > 6.3 × 10-8 > 4.4 × 10-13

Therefore the undissociated (unionised) molecule is always the strongest acid, the species in solution that undergoes the greatest degree of dissociation (ionisation).

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pH Calculations for Weak Polyprotic Acids

If the first dissociation (ionisation) constant, Ka1, is orders of magnitude greater than the second dissociation (ionisation) constant, Ka2, then the original undissociated (unionised) acid molecule contributes the vast majority of the protons in the solution,

Example for diprotic acid H2A:
Stage 1: H2A(aq) H+(aq) + HA-(aq)
Ka1 = [H+][HA-]

[H2A]
> 100 × Ka2
Stage 2: HA-(aq) H+(aq) + A2-(aq)
Ka2 = [H+][A2-]

[HA-]
< 0.01 × Ka1

so the concentration of protons in the acid can be approximated as:
[H+] ≈
Ka1[H2A]

[HA-]

and the pH of the solution can be approximated as:
pH = -log10[H+] ≈ -log10(
Ka1[H2A]

)
[HA-]

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Example: Carbonic Acid

Imagine 1 mole of carbonic acid, H2CO3, is added to 1 L of water6.
What species would be present in the aqueous solution?

Carbonic acid, H2CO3, is a diprotic acid, that is, carbonic acid is able to donate 2 protons.

  acid proton + conjugate base
Stage 1: H2CO3 H+ + HCO3-
Stage 2: HCO3- H+ + CO32-

Carbonic acid is said to be a weak acid, it does not completely dissociate (ionise) in the first stage.

  acid proton + conjugate base   Ka at 25oC
Stage 1: H2CO3 H+ + HCO3-   Ka1 = 4.2 x 10-7
Stage 2: HCO3- H+ + CO32-   Ka2 = 5.6 x 10-11

Notice that the value for Ka1, although small, is much greater than the value for Ka2:

Ka1 ≈ 1000 × Ka2

This means that although there will be few ionic species in the solution, most of the species in solution will be due to the first dissociation equation and that the second dissociation equation will contribute only very, very, small numbers of species.

The following species, due to the dissociation of carbonic acid, will be present in an aqueous solution of carbonic acid:

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Example: pH of Carbonic Acid

Calculate the pH of a 0.50 mol L-1 aqueous solution of carbonic acid at 25°C
given Ka1 = 4.2 × 10-7 and Ka2 = 5.6 × 10-11.

Method 1. Assuming the second acid dissociation (ionisation) stage is negligible:

Ka1 >> Ka2 (Ka1 ≈ 1000 x Ka2)

Therefore the second stage dissociation (ionisation) contributes negligible numbers of protons to the solution and can be ignored.

  1. Write the equation for the first dissociation (ionisation) of carbonic acid in water:

    H2CO3(aq) H+(aq) + HCO3-(aq)   Ka1 = [H+(aq)][HCO3-(aq)]

    [H2CO3(aq)]
    = 4.2 x 10-7

  2. Write expressions for the concentration of each species in solution after the first dissociation:

    (A R.I.C.E. Table is a common way to set out this information)

    R.I.C.E. Table
    Reaction: H2CO3(aq) H+(aq) + HCO3-(aq)
    Initial concentration:
    mol L-1
    0.50   0   0
    Change in concentration:
    mol L-1
    −X   +X   +X
    Equilibrium concentration:
    mol L-1
    0.50 - X   X   X
    Assumption8: X is much, much smaller than 0.50
    So 0.50 - X ≈ 0.50
    approximate equilibrium concentration
    (mol L-1)
    0.50   X   X

  3. Calculate the concentration of hydrogen ions in solution

    Because we are ignoring the second dissociation as contributing a negligible number of protons to the solution, the final concentration of protons in the solution is assumed to be X mol L-1.

    Substitute the values for the concentration of each species into the first dissociation (ionisation) equilibrium expression:

    Ka1 = [H+][HCO3-]

    [H2CO3]

    4.2 × 10-7 = [X][X]

    [0.50]

    2.1 × 10-7 = [X]2

    [X] = 4.6 × 10-4 mol L-1

    [H+] = [X] = 4.6 × 10-4 mol L-1

  4. Calculate the pH of the solution:

    [H+] = 4.6 × 10-4 mol L-1 (calculated above)

    pH = -log10[H+] = -log10[4.6 × 10-4 mol L-1] = 3.3

Method 2. Assume the contribution of the second acid dissociation (ionisation) cannot be ignored

Calculate the pH of a 0.50 mol L-1 aqueous solution of carbonic acid at 25°C
given Ka1 = 4.2 × 10-7 and Ka2 = 5.6 × 10-11.

  1. Write the equation for the first dissociation (ionisation) of carbonic acid in water:

    H2CO3(aq) H+(aq) + HCO3-(aq)   Ka1 = [H+(aq)][HCO3-(aq)]

    [H2CO3(aq)]
    = 4.2 × 10-7

  2. Write expressions for the concentration of each species in solution after the first dissociation:

    (A R.I.C.E. Table is a common way to set out this information)

    R.I.C.E. Table
    Reaction: H2CO3 H+ + HCO3-
    Initial concentration:
    mol L-1
    0.50   0   0
    Change in concentration:
    mol L-1
    −X   +X   +X
    Equilibrium concentration:
    mol L-1
    0.50 - X   X   X
    Assumption: X is much, much smaller than 0.50
    So 0.50 - X ≈ 0.50
    approximate equilibrium concentration
    (mol L-1)
    0.50   X   X

  3. Calculate the concentration of all species in solution

    Substitute the values for the concentration of each species into the first dissociation (ionisation) equilibrium expression:

    Ka1 = [H+][HCO3-]

    [H2CO3]

    4.2 × 10-7 = [X][X]

    [0.50]

    2.1 × 10-7 = [X]2

    [X] = 4.6 × 10-4 mol L-1

    [H+] = [HCO3-] = [X] = 4.6 × 10-4 mol L-1
    [HCO3] = 0.50 - X = 0.50 - 4.6 × 10-4 = 0.499954 mol L-1

  4. Write the equation for the second dissociation (ionisation) of carbonic acid in water:

    HCO3- H+ + CO32-   Ka2 = [H+][CO32-]

    [HCO3-]
    = 5.6 × 10-11

  5. Write expressions for the concentration of each species in solution after the first dissociation:

    (A R.I.C.E. Table is a common way to set out this information)

    R.I.C.E. Table
    Reaction: HCO3- H+ + CO32-
    Inital concentration
    (before dissociation)
    (mol L-1)
    4.6 x 10-4   4.6 x 10-4   0
    Change in concentration:
    mol L-1
    −Y   +Y   +Y
    Equilibrium concentration
    (after dissociation)
    (mol L-1)
    4.6 x 10-4 - Y   4.6 x 10-4 + Y   Y

  6. Substitute the values for the concentration of each species into the second acid dissociation (ionisation) constant expression:

    Ka2 = [H+][CO32-]

    [HCO3-]

    5.6 × 10-11 = [4.6 × 10-4 + Y][Y]

    [4.6 × 10-4 - Y]

  7. Calculate the concentration of each species in the solution:
    5.6 × 10-11 = [4.6 × 10-4 + Y][Y]

    [4.6 × 10-4 - Y]
    5.6 × 10-11[4.6 × 10-4 - Y] = [4.6 × 10-4 + Y][Y]
    2.576 × 10-14 - 5.6 × 10-11Y = 4.6 × 10-4Y + Y2
    0 = Y2 + 4.6 × 10-4Y - 2.576 × 10-14

    Y = [-(4.6 × 10-4) ± √(4.6 × 10-4)2 - 4(1)(-2.576 × 10-14) ]/ 2(1) = 0
    Y = 0 because so very little, practically none, of the HCO3- dissociates.
    Substitute this value for Y back into the expressions for the final concentration of each species:

    R.I.C.E. Table
    Reaction: HCO3-(aq) H+(aq) + CO32-
    Inital concentration
    (before dissociation)
    (mol L-1)
    4.6 × 10-4   4.6 × 10-4   0
    Change in concentration:
    mol L-1
    −0   +0   +0
    Equilibrium concentration
    (after dissociation)
    (mol L-1)
    4.6 × 10-4   4.6 × 10-4   0

  8. Calculate the pH of the carbonic acid solution:

    [H+] in the carbonic acid solution = [H+] present after the second stage of the dissociation = 4.6 × 10-4 mol L-1 (as calculated above)

    pH = -log10[H+] = -log10[4.6 × 10-4] = 3.3

3.Compare the two pH calculations for 0.50 mol L-1 carbonic acid above:


The results of the two pH calculations are identical.
The assumption that the second dissociation contributes negligible amounts of protons to the solution was a good one.

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1. If Ka1 is not orders of magnitude greater than Ka2 then the contribution to [H+] by the second dissociation (ionisation) cannot be ignored in this way.

2. mono from the Greek for alone, monochrome = one colour, monolith = one block of stone, monologue = one person speaks, monatomic = one atom, monomer = a single unit, monovalent = valency of 1, monoxide = 1 oxygen atom

3. poly from Greek for many, polychrome = many colours, polygon = many sided figure, polyatomic ion = ion made up of many atoms, polymer = many repeated units

4. di from Greek for 2, dichromatic = 2 colours, dichotomy = division into 2 parts, diatomic molecule = molecule consisting of 2 atoms, divalent = valency of 2, dioxide = 2 oxygen atoms

5. tri from Greek for 3, triangle = shape with 3 angles, tridentate = 3 teeth, trivalent = valency of 3, trioxide = 3 oxygen atoms

6. Carbonic acid, H2CO3, can not be isolated as a pure substance so we couldn't, in practice, add 1 mole of it to water.
Carbonic acid is used as an example here because school students are usually expected to be aware of the carbonic acid equilibrium in water, and especially of its buffering capacity in natural systems.

7. HCO3- will be named as hydrogen carbonate ion in organic chemistry, but is more likely to be named as hydrogencarbonate ion in inorganic chemistry.

8. If we do not make this assumption we must solve the quadratic equation:

4.2 × 10-7 = [X][X]

[0.50 - X]

4.2 × 10-7 [0.50 - X] = [X][X]

2.1 × 10-7 - 4.2 x 10-7X = X2

X2 + 4.2 × 10-7X - 2.1 × 10-7 = 0

X = [-(4.2 × 10-7) ± √(4.2 × 10-7)2 - 4(1)(-2.1 × 10-7) ]/ 2(1)
= [-4.2 × 10-7 ± 9.16 × 10-4] / 2
= 4.58 × 10-4
(ignore negative values for concentration)