I need to write an original procedure with exact numbers for creating a buffer:

Create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a $\pu{pH}$ of exactly $3.75$.

A $\pu{50 mL}$ sample of your buffered solution has to be able to withstand the addition of $\pu{25.00 mL}$ of $\pu{0.10 M}$ $\ce{NaOH}$ solution.

"Withstand" here is defined as less "than $0.5$ change in $\mathrm{pH}$."

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

We are given $\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid.

Where I've got so far:

I took the $\mathrm{p}K_\mathrm{a}$ of acetic acid $(4.75)$ and the desired $\mathrm{pH}$ $(3.75)$ and put it into the Henderson-Hasselbalch equation:

$$3.75 = 4.75 + \log\frac{[\text{salt}]}{[\text{acid}]}$$

$$\log\frac{[\text{salt}]}{[\text{acid}]} = -1$$


$$\frac{[\text{salt}]}{[\text{acid}]} = 0.10,$$

which means we want the ratio $[\ce{CH3COONa}]:[\ce{CH3COOH}]$ to be $0.10$.

I think we can prepare a solution containing $\pu{0.10 M}$ $\ce{CH3COONa}$ and $\pu{1.00 M}$ $\ce{CH3COOH}.$ But with what we're given $(\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid), I don't know how to proceed and find exact numbers for a procedure.

  • $\begingroup$ It seems to me as almost exact copy of a recent question $\endgroup$ – Poutnik May 31 '19 at 16:39
  • $\begingroup$ chemistry.stackexchange.com/q/115839/35806 $\endgroup$ – Poutnik May 31 '19 at 16:45
  • $\begingroup$ And as requirements here are even more strict than there, neither here the resolution exists. The withstanding and breaking requirements are even more contradictory than before. Other buffer system must be used, with maximum capacity near pH=3.75. Or perhaps even lower. $\endgroup$ – Poutnik May 31 '19 at 16:51

Initial note

The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing.

As consequence, its "breaking" capacity is too big, compared to "withstanding capacity".

General analysis

Let there is given the initial $\mathrm{pH_{ini}}$ and the acid dissociation constant $\mathrm{p}K_\mathrm{a}$

Let there there is given "breaking volume" of $\ce{NaOH}$ $V_\mathrm{br}$, where the buffer losses it's buffering ability.

Let there is given "withstanding volume" $V_\mathrm{ws}$ when $\mathrm{pH}$ changes by value $x$.


$$\mathrm{pH_{ini}}=\mathrm{p}K_\mathrm{a} + \log {R}$$


$$R=\frac{ [\ce{A-}]_\mathrm{init}} {[\ce{HA}]_\mathrm{init}}=10^{\mathrm{pH_{ini}}-\mathrm{p}K_\mathrm{a}}$$

$$\mathrm{pH_{ini}}+x=\mathrm{p}K_\mathrm{a} +\log \frac {R+V_\mathrm{ws}/V_\mathrm{br}\cdot (1-R)}{1-V_\mathrm{ws}/V_\mathrm{br}}$$

If we consider $$R_1=10^{\mathrm{pH_{ini}}+x-\mathrm{p}K_\mathrm{a}} $$ $$V_\mathrm{r}=V_\mathrm{ws}/V_\mathrm{br}$$

then $$R_1=\frac {R+V_\mathrm{r}\cdot (1-R)}{1-V_\mathrm{r}}$$

$$V_\mathrm{r}=\frac {R_1-R}{R_1-R+1}$$


The desired $$V_\mathrm{r}\ge \frac{25}{35}=0.714$$

Acetic acid case

If we consider acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ then $$R=10^{3.75-4.75}=0.1$$

If the change should be $\Delta \mathrm{pH}=0.5$:


$$V_\mathrm{r} =\frac{0.316-0.1}{0.316-0.1+1}=0.177\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:


$$V_\mathrm{r} =\frac{1-0.1}{1-0.1+1}=0.475 \lt 0.714 $$

Formic acid case

If we consider as a replacement e.g. formic acid with $\mathrm{p}K_\mathrm{a}=3.75$ then $$R=10^{3.75-3.75}=1$$

If the change should be $\Delta \mathrm{pH}=0.5$:


$$V_\mathrm{r} =\frac{3.16-1}{3.16-1+1}=0.684\lt 0.714$$

If the change should be $\Delta \mathrm{pH}=1.0$:


$$V_\mathrm{r} =\frac{10-1}{10-1+1}=0.9 \gt 0.714$$


  • Acetic acid is unuseable for making a buffer to fit your requirements.
  • Formic acid can be used to make a buffer with "withstanding" parametr $\Delta~ \text{pH} \lt +1.0$, seen in the other question.
  • Formic acid very tighly missed the "withstanding" parametr $\Delta~ \text{pH} \lt +0.5$, so seems about usable.
  • Even better results could be achieved with an acid with $\mathrm{p}K_\mathrm{a} \lt 3.75$ .
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