I need to create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a pH of exactly $3.75$.

A $\pu{50 mL}$ sample of your buffered solution will have to be able to withstand the addition of $\pu{25.0 mL}$ of $\pu{0.100 M}$ $\ce{NaOH}$ solution.

The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.

Here is what I have so far:

$$\mathrm{pH}= 3.75 \implies [\ce{H+}] = \pu{10^{-3.75} M} = \pu{1.78 \times 10^-4 M}$$

\begin{align} \ce{NaCH3COO &-> Na+ + CH3COO-} &&\text{(common ion effect)}\\ \ce{CH3COOH &<=> H+ + CH3COO-} \end{align}

\begin{align} K_\mathrm{a} &=\frac{[\ce{H+}][\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\ 1.8\times{10^{-5}} &=\frac{(1.78\times{10^{-4}})[\ce{CH3COO-}]} { [\ce{CH3COOH}]}\\ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &=\frac{(1.8\times{10^{-5}})}{(1.78\times{10^{-4}})}\\ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 0.1011 : 1~\text{(ratio)} \end{align}

I found the ratio of $[\ce{NaCH3COO-}]$ to $[\ce{CH3COOH}]$, but I don't know what to do from here.

EDIT: "Break": The point where there's no acid left and the solution stops acting as a buffer. "Withstand": meaning that there can be a $\mathrm{pH}$ change of at most $1.0$. Thanks, guys!

  • $\begingroup$ The "withstand" requirement is not quantified. The "break" I understand as all acid is spent. $\endgroup$
    – Poutnik
    May 23, 2019 at 6:08
  • 1
    $\begingroup$ We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. See Homework $\endgroup$
    – Poutnik
    May 23, 2019 at 6:09
  • 1
    $\begingroup$ Are you familiar with the Henderson-Hasselbalch equation? If you are not, this may help khanacademy.org/science/chemistry/acid-base-equilibrium/… $\endgroup$
    – Waylander
    May 23, 2019 at 7:19
  • $\begingroup$ Can you give us the whole problem exactly as it was stated to you? Then give us your thoughts on how to solve it. $\endgroup$
    – MaxW
    May 23, 2019 at 13:58
  • 1
    $\begingroup$ "withstand" and "break" are not precise terms. How much tolerance for pH are we talking about here? A change of pH of no more than 0.1 unit? Perhaps your instructor has defined these terms quantitatively. $\endgroup$
    – iad22agp
    May 23, 2019 at 23:00

1 Answer 1


In my understanding, there is considered $$\mathrm{p}K_\mathrm{a}=4.75$$ and therefore $\mathrm{pH}=3.75$ was chosen for easy calculation, as:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log \frac {c_{\ce{A-}}}{c_{\ce{HA}}}$$

and therefore $\frac {c_{\ce{A-}}}{c_{\ce{HA}}}=0.1$

For the buffer change to $\mathrm{pH}=4.75$, at least $\pu{25 ml 0.1 M NaOH}$ has to be spent. ("withstand")

$$\frac {c_{ws,\ce{A-}}}{c_{ws,\ce{HA}}}=1$$

Calculating molar amounts to be equal for both acetic acid and acetate: $$0.1\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}+c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}\\=c_{\ce{HA}}\cdot V_\mathrm{buf}-c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}$$

From that, we calculate minimal concentration of acetic acid in buffer:

$$2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}=0.9\cdot c_{\ce{HA}}\cdot V_\mathrm{buf}$$

$$c_{\ce{HA}}=\frac{2\cdot c_\mathrm{NaOH}\cdot V_\mathrm{NaOH}}{0.9\cdot V_\mathrm{buf}}=\frac{2\cdot 0.1\cdot 0.025}{0.9\cdot 0.05}=0.005/0.045=1/9 \mathrm{~ mol/l}$$

$$c_{\ce{HA},min}=\frac19 \mathrm{~ mol/l}$$ $$c_{\ce{A-},min}=0.1 \cdot \frac19=\frac1{90} \mathrm{~ mol/l}$$

All acetic acid in $\pu{50 mL}$ of the buffer must react with not more than $35~\text{ml}$ of $0.1~\mathrm{M~}\ce{NaOH}$.

$$\begin{align} n_{\ce{NaOH}}&\lt 3.5\mathrm{~mmol}\\ n_{\ce{CH3COOH}}&\lt 3.5\mathrm{~mmol}\\ n_{\ce{CH3COONa}}&\lt 0.35\mathrm{~mmol}\\ \end{align}$$

As the latter 2 amounts have to be present in $\pu{50 mL}$ of the buffer:

$$\begin{align} c_{\ce{CH3COOH}}&\lt \pu{70 mmol/L}\\ c_{\ce{CH3COONa}}&\lt \pu{7 mmol/L}\\ \end{align}$$

Summary: Concentration of acetic acid in the buffer must be at least $\frac19=0.1111\pu{ mol/L}$ and not more then $\pu{0.07 mol/L}$.

Such requirements is impossible to meet and such a buffer is not possible to prepare.

By other words, the buffer cannot have at the same time so big buffer ("withstand") capacity and so low "break" capacity.

If usage of acetate buffer was not mandatory, the goal would be achievable with buffers with pKa closer to 3.75 than the acetic acid has. In such a case, the buffer capacity would be decreasing and not increasing between these pH values and the final break would be faster.


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