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pH After Mixing Aqueous Solutions of Weak Acids Chemistry Introductory Tutorial

Key Concepts

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Mixing Weak Acids: An Introduction to Simultaneous Equilibria

An acid, HA(aq), is weak because it only partially dissociates in water: (3)

Arrhenius: HA(aq) H+(aq) + A-(aq)   Ka = [H+(aq)][A-(aq)]
[HA(aq)]
Brønsted-Lowry: HA(aq) + H2O(l) H3O+(aq) + A-(aq)   Ka = [H3O+(aq)][A-(aq)]
[HA(aq)]

The larger the value of the acid dissociation constant (acid ionisation constant), Ka, the greater the degree of acid dissociation and the greater the concentration of hydrogen ions in solution.
The greater the concentration of hydrogen ions, the lower the pH.

But what if you add one weak acid to a different weak acid? What happens to the concentration of hydrogen ions then? How do you calculate the pH of the resultant solution?

Consider an aqueous solution of acetic acid (ethanoic acid), CH3COOH(aq), as an example of a weak acid.
Firstly, acetic acid is a weak acid so it partially dissociates in water. At 25°C, Ka = 1.8×10-5

CH3COOH(aq) H+(aq) + CH3COO-(aq)   Ka = [H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)]
= 1.8×10-5
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq)   Ka = [H3O+(aq)][CH3COO-(aq)]
[CH3COOH(aq)]
= 1.8×10-5

But water molecules which make up the solvent for the acetic acid solution also undergo dissociation to produce hydrogen ions and hydroxide ions. At 25°C, Kw = 1.0×10-14

H2O(l) H+(aq) +OH-(aq)   Kw = [H+(aq)][OH-(aq)] = 1.0×10-14
2H2O(l) H3O+(aq) + OH-(aq)   Kw = [H3O+(aq)][OH-(aq)] = 1.0×10-14

This is one example of simultaneous equilibria because both of these dissociation reactions are occurring at the same time in the same solution.
However, because the value for the acid dissociation constant for acetic acid (Ka) is many orders of magnitude greater than the value for the self-dissociation of water (Kw), we can often ignore the contribution of the dissociation of water to the concentration of hydrogen ions in the final solution (4):

Ka >> Kw then [H+(aq)] is due to the dissociation of the acid.

Ka >> Kw then [H3O+(aq)] is due to the dissociation of the acid.

By extension, if we then added another weak acid to the acetic acid, for example, we could add hydrocyanic acid (HCN(aq)) for which Ka = 6.3×10-10 at 25°C, we would have a solution in which the conditions for 3 simultaneous equilibria must be met:

(i) CH3COOH(aq) ⇋ H+(aq) + CH3COO-(aq) Ka = 1.8×10-5
(ii) HCN(aq) ⇋ H+(aq) + CN-(aq) Ka = 6.3×10-10
(iii) H2O(l) ⇋ H+(aq) + OH-(aq) Kw = 1.0×10-14

But, since Ka for the dissociation of acetic acid is orders of magnitude greater than Ka for the dissociation hydrocyanic acid, which is much greater than Kw, we can say that the dissociation of acetic acid contributes almost all of the hydrogen ions in solution, so we use only the value of its dissociation constant when calculating [H+(aq)].

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Worked Example of Calculating the pH of a Solution After Mixing Two Weak Acids

Imagine we add 100.0 mL 0.200 mol L-1 HCN(aq) to 100.0 mL 0.200 mol L-1 CH3COOH(aq).

In the discussion above we wrote the equations for the 3 simultaneous equilibria:

(i) CH3COOH(aq) ⇋ H+(aq) + CH3COO-(aq) Ka = 1.8×10-5
(ii) HCN(aq) ⇋ H+(aq) + CN-(aq) Ka = 6.3×10-10
(iii) H2O(l) ⇋ H+(aq) + OH-(aq) Kw = 1.0×10-14

and noted that since Ka(CH3COOH(aq)) >> Ka(HCN(aq)) >> Kw we can ignore the contributions of the dissociation of HCN(aq) and H2O(l) to the final concentration of H+(aq) in solution.

In effect we have diluted our original 0.200 mol L-1 acetic solution since its original volume was 100.0 mL, but now the solution has a volume of 100.0 mL + 100.0 mL = 200.0 mL (assuming additivity of volumes).

We can calculate the concentration of this diluted acetic acid solution:

c1 = 0.200 mol L-1

V1 = 100.0 mL/1000 mL/L = 0.100 L

c2 = ? mol L-1

V2 = 200.0 mL/1000 mL/L = 0.200 L

c1V1 = c2V2

0.200 × 0.100 = c2 × 0.200

0.100 mol L-1 = c2

Now we can use this as the initial concentration of acetic acid before it dissociates which will allow us to calculate [H+(aq)] at equilibrium.

First we set-up the R.I.C.E. Table as shown below:

Let x represent the change in concentration of species.

As a result of the weak acid dissociation:

⚛ concentration of reactant acid molecules decreases

⚛ concentration of product ions increases

Reaction: CH3COOH(aq) H+(aq) + CH3COO-(aq)
Initial concentration:
mol L-1
0.100   0   0
Change in concentration:
mol L-1
x   +x   +x
Equilibrium concentration:
mol L-1
0.100 −x ≈ 0.100
( assume x << 0.100)
  x   x

Next, we use the value of Ka for acetic acid to calculate x and hence [H+(aq)]:

Ka = [H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)]
= 1.8×10-5
  [x][x]
[0.100]
= 1.8×10-5
  x2 = 1.8×10-6
  x = 1.3×10-3
  [H+(aq)] = 1.3×10-3 mol L-1

And now we can calculate the pH of the resultant solution:

pH = −log10[H+(aq)] = −log10[1.3×10-3] = 2.9

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Footnotes:

(1) We are also assuming that the weak acids we are mixing do not react with each other.

(2) We are also assuming that the acid dissociation constants for both acids are greater than the value for the self-dissoication of water:
Ka(HA) >> Ka(HZ) >> Kw

(3) Dissociation is an Arrhenius acid-base concept, that is, an acid (or a base) dissociates (breaks apart) in water to produce ions (the first equation given). Ionisation, or ionization, is often used instead of dissociation in this context, which students can find confusing because ionisation refers to the removal of an electron from an atom using energy, and, ionising radiation is radiation that is able to knock electrons off atoms. However, you will find the terms "acid dissociation", "acid ionisation" and "acid ionization" used synonymously.
An Arrhenius acid is a strong acid if it completely dissociates, an acid is weak if it only partially dissociates.
The Brønsted-Lowry definition of an acid is that an acid is a species that donates a proton to another species (the second equation given). So an acid can donate a proton to solvent water molecules to produce an oxidanium ion, H3O+ (acceptable alternative non-systematic IUPAC name oxonium, and in old textbooks, hydronium). The reaction between an acid molecule and the solvent is therefore seen as a proton-transfer reaction. For a strong acid, the equilibrium position lies very far to the right. For a weak acid, the equilibrium position lies far to the left.
You can use either the Arrhenius definition or the Brønsted-Lowry definition of an acid when writing the expression for Ka, and you can use either [H+] or [H3O+] in calculations.

(4) If the weak acid solution is extremely dilute, then the concentration of hydrogen ions approximates the concentration of hydrogen ions in water!
(see pH of Aqueous Solution of Weak Acid After Dilution Tutorial)