H+ concentration of the mixture of a strong and weak acid

Question

I reviewed the complete solution (hand-written) to the problem at Calculation of the pH of a mixture of a strong acid and weak acid.

My question is why it is okay to apply the K1 value for the weak acid to the mixture of the weak acid and strong acid. If K1 is the equilibrium constant for the weak acid, it seems like it should not apply to the mixture, only to the weak acid. If K1 only applies to the weak acid, wouldn't the equation for K1 be: K1=(x^2)/(C1-x)? What am I missing?

Answer 1

In the handwritten answer, $K_1$ is the equilibrium constant for the reaction $\ce{HA <=> H+ + A-}$, where $\ce{HA}$ is a weak acid.

The equilibrium equation

$$K_1 = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$$

is true no matter where the $\ce{H+}$ and the $\ce{A-}$ come from. For example, adding a completely separate source of $\ce{H+}$ will shift the equilibrium of $\ce{HA <=> H+ + A-}$ to the left. This reflects the fact that every time $\ce{H+}$ and $\ce{A-}$ collide, there is a chance that they combine to form $\ce{HA}$. Adding more $\ce{H+}$ (or more $\ce{A-}$) from any source will therefore increase the concentration of $\ce{HA}$.

In this case, all the $\ce{A-}$ comes from the weak acid, but the $\ce{H+}$ can come from the weak acid or from the strong acid. (It can also come from water; this possibility was disregarded in the answer you cited, an approximation which is acceptable as long as $K_1$ is substantially larger than $K_{water}$.)

The handwritten answer introduces the following notation:

• $x$ is the equilibrum value of $\ce{[A-]}$
• $c_1$ is the initial concentration of $\ce{HA}$ (that is, the concentration that would be present if it did not react)
• $c_2$ is the initial concentration of $\ce{HB}$, the strong acid (again, this is the concentration of $\ce{HB}$ that would be present if it didn't dissociate at all).

Using this notation:

1. $\ce{[A-]} = x$ by definition.
2. $\ce{[HA]} = c_1-x$ because $c_1$ is the original amount of $\ce{HA}$ and $x$ is the amount that dissociated.
3. $\ce{[H+]} = x+c2$ because $\ce{H+}$ comes from two sources. $x$ is the amount produced by the dissociation of $\ce{HA}$. $c_2$ is the amount produced by the dissociation of $\ce{HB}$, assuming that $\ce{HB}$ fully dissociates, which is a good approximation because it's a strong acid.

Therefore $$K_1 = \frac{\ce{[H+][A-]}}{\ce{[HA]}} = \frac{(x+c_2)x}{c1-x}.$$