How to create a buffer using CH3COOH and CH3COONa that has a pH of exactly 3.75?

Question

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!

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.