Suppose I want to make a buffer solution in any of the three ways possible with dihydrogen phosphate as my weak acid. In the first way, a certain ratio of the weak acid and its conjugate base are added at once to form the buffer. Why does this not return to the equilibrium described by its Ka value instead of forming a stable buffer? With a pka of 7.2, shouldn't the mixture shift to the weak acid to reestablish equilibrium. The other two ways involve adding a strong acid or base to a weak base or weak acid respectively, but I still see the same issue. Say a strong base is added to a solution of the dihydrogen phosphate and the same moles added react and form an equal number of moles of the conjugate base. After this is complete, why doesn't equilibrium get restored. Is the weak acid/base equilibrium out-competed by the seperate strong base/weak acid to weak acid/weak base equilibrium?

• Can you explain why you think the buffered solutions are not at equilibrium? – a-cyclohexane-molecule Aug 29 '18 at 2:20
• @a-cyclohexane-molecule Well say you're using a weak acid (not so weak that the conjugate base is especially strong). Only very few molecules of this acid should dissociate. In preparing a buffer with a weak acid and its conjugate weak base at the same time, isn't the solution suddenly not in equilibrium because there are too many molecules of base for every one of the weak acid? Why doesn't the base just convert itself to back to the acid to restore the equilibrium ratio that would exist if it were just the weak acid isolated in the solution? – Elmer Aug 29 '18 at 5:58
• In fact, equilibrium does restore, and that pretty fast. – Ivan Neretin Aug 29 '18 at 7:11
• At equilibrium, the ratio $\ce{[HA]}/\ce{[A-]}$ is not necessarily constant. The ratio $\ce{[HA]}/(\ce{[H+]}\ce{[A-]})$, however, is. – a-cyclohexane-molecule Aug 29 '18 at 12:08
• @a-cyclohexane-molecule Gotcha. So how is it that it can be relied on that the [H+] will rise to the necessary amount for the desired pH after the weak acid and weak base are added? Shouldn't [H+] lower itself after it's added since the amounts of the WA and WB don't change. That way the ratio of the Ka value remains small. I feel like I must be missing something big here – Elmer Aug 30 '18 at 3:16

Quantitatively, the equilibrium expression can be rearranged into the Henderson-Hasselbalch equation $$\text{pH} = \text{pKa} + \log\left(\frac{\ce{[A-]}}{\ce{[HA]}}\right) \approx \text{pKa} + \log\left(\frac{0.1+x}{0.1-x}\right) \approx \text{pKa} + \log\left(1+20x\right) \approx \text{pKa} + 20x \approx \text{pKa},$$ where $x$ represents the concentration of HA that has dissociated, assumed small. The third equality follows from a Taylor expansion of $0.1-x$, and the fourth from a Taylor expansion of $\log(1+20x)$. We can check whether this quantity is indeed small by solving the equilibrium expression for $x$ exactly; intuitively, it should be small because the difference in proton concentration at a pH of 5 and at a pH of 7 is small relative to the concentration of the acid.