# pH of an acetate buffer after dilution

100 mL of a 0.1 mM buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?

The pKa value for acetic acid is given as 4.75.

The answer provided in my textbook is pH = 5.53. The closest that I can get to that is pH = 5.507. I know that's an almost insignificant difference but I'm just confused how they got to their answer, so if anyone can give me some suggestions that would be awesome. I'll outline my approach below:

Using the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] in the 100ml: $$10^{5-4.75} = \frac{[A^-]}{[HA]}$$

$$[A^-] = \left(\frac{10^{0.25}}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 6.4{\times}10^{-5}M$$

$$[HA] = \left(\frac{1}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 3.6{\times}10^{-5}M$$

The addition of 900ml of water to 100ml of buffer is a 10x dilution:

$$[A^-] = 6.4{\times}10^{-5} \times 0.1 = 6.4{\times}10^{-6}M$$

$$[HA] = 3.6{\times}10^{-5} \times 0.1 = 3.6{\times}10^{-6}M$$

$$\ce{[H+]} = 10^{-5} \times 0.1 = 10^{-6}M$$

ICE table to determine equilibrium concentrations after dilution:

$$\begin{array}{c|c@{hello}c@{}c@{}c@{}c@{}c@{}c} & [CH_3 COOH] & + &[H_2 O] & {}\rightleftharpoons{} & [H_3 O^+] & + &[CH_3 COO^-] \\ \hline I & 3.6 \times 10^{-6} && - && 10^{-6} && 6.4 \times 10^{-6} \\ C & -x && - && +x && +x \\ E & 3.6 \times 10^{-6} -x && - && 10^{-6} +x && 6.4 \times 10^{-6} +x \\ \hline \end{array}$$

$$K_a = \frac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$ Then substituting in Ka and the equilibrium values from the ICE table: $$(10^{-4.75}) = \frac{(10^{-6}+x)(6.4 \times 10^{-6} + x)}{(3.6 \times 10^{-6} -x)}$$

$$x^2 + 2.518 \times 10^{-5}x -5.762 \times 10^{-11} = 0$$

$$x = 2.111 \times 10^{-6}$$ $$[H^+] = 10^{-6} + 2.111 \times 10^{-6} = 3.111 \times 10^{-6}$$

$$pH = -log_{10}(3.111 \times 10^{-6}) = 5.507$$

• Your intermediate results have two significant figures. You answer and the textbook answer are the same at two significant figures. If you want to get closer, keep some extra digits in the intermediate results. – Karsten Theis Jul 1 '19 at 20:44
• Calling a solution where the total concentration of acid and base combined is 0.1 mM (or 0.01 mM) a buffer is unconventional, to say the least. As you can see from your calculation, the conjugate acid is no longer a major species. The buffer capacity will be very low, and adding an indicator at concentrations where you can see it will change the pH by a lot. – Karsten Theis Jul 1 '19 at 20:48
• That was one of my first thoughts, but even using a value for x calculated using unsimplified values for [HA] and [A-] throught the calculation results in an almost identical answer: pH = 5.50716. I also agree that the solution isn't exactly a great buffer - but then that depends on what it has to buffer. My problem isn't with the peculiarity of the question, but rather with the approach to finding the pH of a highly diluted weak acid/buffer system. – Will Jul 1 '19 at 21:16
• You might have spent more time on the question than the person how gave the textbook answer... – Karsten Theis Jul 1 '19 at 21:21
• If the initial pH was 5.0 you know that only has one significant figure, right? I hate these sort of stupid problem where they state the problem with few significant figures. If they want 3 significant figures in the answer then the problem should be setup in such a way to support that. // As it is the book's problem solver probably rounded somewhere and as you now know you'll go crazy trying to figure out where. // In the good old days I use a slide rule for about 10 years. 2 significant figures was typical, three if you were careful. – MaxW Jul 2 '19 at 3:26

Depending on your first calculations before dilution: \begin{align} [\ce{CH3COO-}]&=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\ [\ce{Na+}]_\text{(After dilution)}&=\pu{6.4\times{10^{-6} M}}\\ [\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\ [\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-5} M}}~\text {(After dilution)} \end{align} So,you have the the following $$5$$ independent equations in order to specify the unknown concentrations of $$5$$ species as present in an aqueous solution consisting of $$\pu{1\times{10^{-5} M}}$$ buffer solution made from acetic acid and sodium acetate :
\begin{align} 6.4 \times 10^{−6} &= [\ce{Na+}] \tag{1}\\ K_\mathrm{a} &= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = 1 \times 10^{−4.75} \tag{2}\\ K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] = 1 \times 10^{−14} \tag{3}\\ 1\times{10^{-5}} &=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\ [\ce{H+}]+[\ce{Na+}] &= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5} \end{align}