How do I write a procedure for creating a buffer?

Question

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:

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

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

So

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

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.

Answer 1

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 $x$.

Then:

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

where

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

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

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

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

Goal

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 $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-4.75}=0.316$$

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

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

$$R_1=10^{3.75+1.0-4.75}=1$$

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

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 $\Delta \mathrm{pH}=0.5$:

$$R_1=10^{3.75+0.5-3.75}=3.16$$

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

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

$$R_1=10^{3.75+1.0-3.75}=10$$

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

Summary

• 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 $\mathrm{p}K_\mathrm{a} \lt 3.75$ .