equilibrium - Calculation of the pH of a mixture of a strong acid and weak acid

The question is to find out the $\mathrm{pH}$ of a mixture of weak acid and strong acid. My book just states the formula as $$\mathrm{pH}=-\log \frac{C_2+\sqrt{C_2^2+4K_\mathrm{a}C_1}}{2}$$

where $C_1$ is the concentration (in mole/litre) of the weak acid (ionisation constant $K_\mathrm{a}$), and $C_2$ is that of the strong acid.

I understand that the strong acid will dissociate completely and the weak acid will dissociate with an ionisation constant. So the $\ce{H+}$ ion concentration from strong acid will be simply $C_2$ and that from weak acid will be $\sqrt{K_\mathrm{a}C_1}$. But I could not derive the above formula for the $\mathrm{pH}$ using these facts. I need help here.

Answer

Chemical thermodynamics is easy in that it always has enough equations to derive all the variables involved, and guarantees to produce a unique solution. The derivation itself may be not that easy, but that's another story.

In short, we have some weak acid $\ce{HA}$ and the products of its dissociation. Mind you, $\ce{H+}$ comes also from the strong acid, which dissociates completely. Now, the equations:

$$ \ce{[HA] + [A-] = }C_1 \\ \ce{[H+]=[A-]}+C_2 \\ {\ce{[H+][A-]}\over\ce{[HA]}} = K_\mathrm{a} $$

Is the rest clear?

Use the second equation to exclude $\ce{[A-]}$, then use the first to exclude $\ce{[HA]}$, and end up with a quadratic equation on $\ce{H+}$.