How to calculate pH of a solution with two dominant equilibria?

Question

Say you have acid X for which $K_a(X)$ = 1E-7 and acid Y for which $K_a(Y)$ = 0.8E-7. In separate solutions, it would be easy to calculate $[H^+]$ because there would be only one dominant equilibrium. But in solution together, both contribute significantly to $[H^+]$, and as a result, both equilibria must be driven to the left by Le Chatelier's Principle. How do you calculate the pH of such a solution? Since both contribute to $[H^+]$, the resulting $[H^+]$ is not additive, but rather a bit less than that because of suppression.

Assume X and Y are 0.1M and monoprotic.

Answer 1

The first step is to write down the variables of interest, and the equations that relate them. If there exactly as many (non-redundant) equations as variables, then the system might be solvable.

In this case, the variables are:

1. $\ce{[H+]}$
2. $\ce{[X^{-}]}$
3. $\ce{[Y^{-}]}$
4. $\ce{[HX]}$
5. $\ce{[HY]}$
6. $\ce{[OH^{-}]}$

The equations are:

1. water dissociation equilibrium: $K_w=\ce{[H+][OH^{-}]}$
2. HX dissociation equilibrium $K_{a,x}=\frac{\ce{[X^{-}][H+]}}{\ce{[HX]}}$
3. HY dissociation equilbrium $K_{a,y}=\frac{\ce{[Y^{-}][H+]}}{\ce{[HY]}}$
4. charge balance: $\ce{[H+]}=\ce{[OH^{-}] + [X^{-}] + [Y^{-}]}$
5. X mass balance: $\ce{[X^{-}] + [HX]}=\text{constant}=0.1\text{ molar}$
6. Y mass balance: $\ce{[Y^{-}] + [HY]}=\text{constant}=0.1\text{ molar}$

So, "all" you have to do is solve those equations for the six unknown variables. That isn't easy to do with pen and paper, but it can be done. Or you could use a computer to numerically solve the equations since you know all the numbers $K_w$, $K_{a,x}$, $K_{a,y}$, and the two different 0.1 molar total concentrations.

Once you solve the equations for $\ce{[H+]}$, the pH is just calculated from $pH = -\log_{10}\ce{[H+]}$.

Answer 2

Curt's answer is correct, but as a practical matter considering the particular concentrations and ionizaiton constants in the question, we can reduce the problem to 3 equations and 3 unknowns as:

$K_{ax}=\frac{[X-][H+]}{0.1M}$

$K_{ay}=\frac{[Y-][H+]}{0.1M}$

$[H+]=[X-] + [Y-]$

And substituting:

$[H+]= \frac{0.1M~K_{ax}}{[H+]} + \frac{0.1M~K_{ay}}{[H+]}$

$[H+]^2 = 0.1M~K_{ax} + 0.1M~K_{ay}$

$[H+]^2 = 1.8 \times 10^{-8}M^2$

$[H+] = 1.34 \times 10^{-4}M$

$pH = 3.9$

So how accurate are the approximations?

Instead of zero, [OH-] is $7.46 \times 10^{-11}M$

So really $[H+]=[X-] + [Y-] + 7.46 \times 10^{-11}M$

About a 1 part in 1,000,000 error in [H+] from neglecting the autoionization of water. If the concentration of acids was very small instead of 0.1M, and pH closer to 7, this error could be much larger.

The error in approximating [HX] and [HY] as 0.1M is about 1 part in 1,000