A biochemist needs a medium for acid-producing bacteria. The pH of the medium must not change by more than $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{pH}$units for every$\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{mol}$of${\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}$generated by the organisms per litre of medium. A buffer consisting of$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{M}\mathbf{}\mathbf{HA}$and$\mathbf{0}\mathbf{.}\mathbf{10}{\mathbf{MA}}^{\mathbf{-}}$is included in the medium to control its$\mathbf{pH}$. What volume of this buffer must be included in$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{L}$of medium?

The Henderson–Hasselbalch equation $\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}{\mathbf{log}}_{\mathbf{10}}\left(\frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Acid}\right]}\right)$in chemistry and biology connects the role="math" localid="1663269802447" $\mathbf{\text{pH}}$of a weak acid chemical solution to the numerical value of the acid dissociation constant, ${\mathbf{\text{K}}}_{\mathbf{a}}$, and the ratio of the acid and its conjugate base concentrations$\frac{\mathbf{\left[}\mathbf{Base}\mathbf{\right]}}{\mathbf{\left[}\mathbf{Acid}\mathbf{\right]}}$in equilibrium.

data-custom-editor="chemistry" $\mathbf{HA}\mathbf{\left(}\mathbf{acid}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{B}}^{\mathbf{-}}\mathbf{(}\mathbf{base}\mathbf{)}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}$

For the problems with buffer solutions, use Henderson-Hasselbalch equation –

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{{\mathrm{A}}^{-}}{\mathrm{HA}}\right)$

Moles of acid and base can also be used instead of their concentrations.

Rearrange this equation and subtract ${\mathrm{pK}}_{\mathrm{a}}$value from both sides and get the change in$\text{pH}$(only if concentrations of base and acid are equal) –

$\begin{array}{c}\mathrm{\Delta pH}=\log \frac{\mathrm{n}\left(\mathrm{acid}\right)}{\mathrm{n}\left(\mathrm{base}\right)}\\ -0.05=\log \frac{1\mathrm{L}·\mathrm{xM}-0.001\mathrm{mol}}{1\mathrm{L}·\mathrm{xM}+0.001\mathrm{mol}}\\ 0.8913=\frac{\mathrm{x}-0.001\mathrm{mol}}{\mathrm{x}+0.001\mathrm{mol}}\\ \mathrm{x}=0.0174\mathrm{mol}\end{array}$

This is the concentration of acid and base that is required$1$ in of medium. Now calculate the volume of $0.1\text{M}$acid and $0.1\text{M}$base required to create $1\text{L}$of $0.0174\text{M}$buffer –

$\begin{array}{c}{\mathrm{c}}_1{\mathrm{V}}_1={\mathrm{c}}_2{\mathrm{V}}_2\\ 0.1\mathrm{M}\times {\mathrm{V}}_1=0.0174\mathrm{M}\times 1\mathrm{L}\\ {\mathrm{V}}_1=0.174\mathrm{L}=174\mathrm{mL}\end{array}$

Therefore, the value for volume is obtained as ${\mathrm{V}}_1=0.174\mathrm{L}=174\mathrm{mL}$.