# 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.

• can anyone point out which book that provides the stated equation? thank you in advance Jan 21, 2022 at 2:08

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:

$$\ce{[HA] + [A-] = }C_1 \tag1$$ $$\ce{[H+]=[A-]}+C_2 \tag2$$ $${\ce{[H+][A-]}\over\ce{[HA]}} = K_\mathrm{a} \tag3$$

Is the rest clear?

Use the equation $$(2)$$ to exclude $$\ce{[A-]}$$, then use the equation $$(1)$$ to exclude $$\ce{[HA]}$$, and end up with a quadratic equation on $$\ce{H+}$$.

• Well, the first one is just the conservation law: the summary concentration of protonated and deprotonated forms of HA is always the same, for it can't be created or destroyed. The second is charge conservation: for every H+, there is an A- or the other anion. Feb 26, 2017 at 13:58
• Nope. I used A for the anion of the weak acid. The strong acid I didn't identify at all, because why would I? Feb 26, 2017 at 16:33
• @IvanNeretin I derive this formula according to your method :$$\mathrm{pH}=-\log \frac{-(K_a-C_2)+\sqrt{(K_a-C_2)^2+4K_\mathrm{a}(C_1+C_2)}}{2}$$ Aug 18, 2018 at 11:34
• First formula is OK. Second formula is OK if Ka is negligibly low compared to C2, but in that case it can be simplified further. Aug 18, 2018 at 12:03
• Ignore the dissociation of the weak acid altogether. Aug 23, 2018 at 17:09 -The above workup should be pretty easy to understand.

-The only tough task is to understand the approximation in 3rd last step.

• Ivan's answer was sufficient. Apr 19, 2018 at 8:37
• He just gave hints.The above answer is complete workup. Apr 19, 2018 at 8:37
• The answer was accepted which means the OP was able to do the working himself. It's just about solving a quadratic. Apr 19, 2018 at 8:39
• Hey Ritwik, thanks for your answer! Unfortunately though, most of what you wrote is already covered by Ivan's answer. Only a quadratic equation needed to be solved, as Abcd also said. The only new thing I see is a simple statement: "$K_1<C_2$ (magnitude wise)". Perhaps, that was better suited as a comment on Ivan's answer instead? || Also, we support mathjax on our site. You needn't paste photos of math expressions, head over here to learn more. Thanks! Apr 19, 2018 at 9:11

It should be noted that this formula is applicable only if the acid constant of the weak acid is indeed much smaller than the concentration of the H3O+ ions formed in the protolyzation of the strong acid. This is because the formula uses the approximation K1 << C2 in the derivation.

For example, if:

$$K_1 = 5 \cdot 10^{-2},$$

and the

$$C_2 = 0.01 M,$$

then the suggested formula is not valid, which is obviously because the used approximation (K1 << C2) does not hold.