pH After Mixing Aqueous Solutions of Weak Acids Introductory Chemistry Tutorial

Key Concepts

• Different weak acids have different values for their acid dissociation constants (acid ionisation constants), K_a

  Acid A:	HA(aq)	⇋	H+(aq) + A-(aq)		Ka(HA) =	[H+(aq)][A-(aq)] [HA(aq)]
  Acid Z:	HZ(aq)	⇋	H+(aq) + Z-(aq)		Ka(HZ) =	[H+(aq)][Z-(aq)] [HZ(aq)]

  K_a(HA) ≠ K_a(HZ)

• When two weak acids are mixed, each contributes H⁺(aq) (or H₃O⁺(aq)) to the resultant solution.⁽¹⁾
• One of the weak acids is often much weaker than the other, that is, the value of one acid dissociation constant is much smaller than the other:

  If Acid Z, HZ(aq), is a much weaker acid than Acid A, HA(aq),

  Then K_a(HZ(aq)) << K_a(HA(aq))

• When one K_a value is much smaller than the other, we ignore the H⁺(aq) (or H₃O⁺(aq)) contribution from the weaker acid when calculating the pH ⁽²⁾

  For K_a(HZ(aq)) << K_a(HA(aq))

  [H⁺(aq)] is determined by K_a(HA(aq))

  [H+(aq)]

  =

  Ka(HA)[HA(aq)] [A-(aq)]

  Therefore pH is determined by K_a(HA(aq))

  pH = −log₁₀[H⁺(aq)]

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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⁾

The larger the value of the acid dissociation constant (acid ionisation constant), K_a, 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, K_a = 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, K_w = 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 (K_a) is many orders of magnitude greater than the value for the self-dissociation of water (K_w), we can often ignore the contribution of the dissociation of water to the concentration of hydrogen ions in the final solution ⁽⁴⁾:

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

K_a >> K_w 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 K_a = 6.3×10^-10 at 25°C, we would have a solution in which the conditions for 3 simultaneous equilibria must be met:

But, since K_a for the dissociation of acetic acid is orders of magnitude greater than K_a for the dissociation hydrocyanic acid, which is much greater than K_w, 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)].

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:

and noted that since K_a(CH₃COOH(aq)) >> K_a(HCN(aq)) >> K_w 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

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

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

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