Q19.143 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

Volume of the buffer that must be included in $1.0 \mathrm{L}$ of medium is ${\mathrm{V}}_1=174\mathrm{mL}$.

Step 1: Concept Introduction.

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 $\mathbf{\text{pH}}$ 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" $\mathbf{HA}\mathbf{\left(}\mathbf{acid}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{B}}^{\mathbf{-}}\mathbf{(}\mathbf{base}\mathbf{)}\mathbf{+}{\mathbf{H}}^{\mathbf{+}}$

Step 2: Henderson-Hasselbalch Equation.

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) –

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

Step 3: Volume of Buffer.

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 ${\mathrm{V}}_1= 0.174\mathrm{L} = 174\mathrm{mL}$.