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.


2 Answers 2


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 $pH = -\log_{10}\ce{[H+]}$.


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:



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

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

  • $\begingroup$ I like this answer, but could you spell out the assumptions you are making? E.g., compared to Curt's answer you ignore the ionization of water, and you fill in 0.1 M for the concentration of the acids when we know they are ionizing as well. I think those are what gives people problems when tackling these types of problems. $\endgroup$
    – jerepierre
    Feb 27, 2015 at 15:59
  • $\begingroup$ @jerepierre I'll try to add more explanation, but the main sources of error would be approximating pH as -log[H+], instead of -log (H+ activity), approximating that there exists a pKa that is a product of concentrations only (rather than activities or concentrations and activity coefficients) and is independent of pH; and to a lesser extent neglecting the concentration of water in the equilibrium. The after considering all those approximations that would apply to both what Curt and I have done, there are my approximations. $\endgroup$
    – DavePhD
    Feb 27, 2015 at 16:18

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