Question

What mass of sodium benzoate should be added to 140 mL of a 0.15M benzoic acid solution to obtain a buffer with a pH of 4.25?

Answer 1

Hint: A buffer is a solution that can withstand changes in pH caused by the addition of acidic or essential ingredients. It may neutralise minor quantities of added acid or base, allowing the pH of the solution to remain comparatively unchanged. This is critical for processes and/or reactions that necessitate complex and stable pH ranges.

Complete answer:
When the proportions of the acid and the conjugate base are the same, i.e. when the acid is 50% dissociated, the pH of the solution equals the pKa of the acid, according to the Henderson-Hasselbach equation. This interaction is used to experimentally establish the pKa of compounds.
 $ {\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + {\log _{10}}\left( {\dfrac{{\left[ {{A^ - }} \right]}}{{[HA]}}} \right) $
Benzoic acid and sodium benzoate are combined to make the buffer solution.
Calculate the amount of moles of salt used to generate the buffer solution with the optimal pH using the Handerson-Hasselbalch equation.
Tables at the end of most chemistry textbooks display the pKa values for the most important weak acids.
Benzoic acid has a pKa value of 4.20.
 $ \begin{array}{*{20}{l}}
  {{n_{{\text{Salt }}}} = ?} \\
  {{n_{{\text{Acid }}}} = {C_M} \times V} \\
  {{n_{{\text{Acid }}}} = 0.15\dfrac{{{\text{ mol}}{\text{. }}}}{L} \times 0.1400L} \\
  {{n_{{\text{Acid }}}} = 0.0210{\text{ mol}}{\text{. }}}
\end{array} $
Now we solve for the unknown
 $ \begin{array}{*{20}{l}}
{pH = p{K_a} + \log \left( {\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}}} \right)}\\
{4.25 = 4.20 + \log \left( {\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}}} \right)}\\
{\log \left( {\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}}} \right) = 4.25 - 4.20}\\
{\log \left( {\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}}} \right) = 0.05}\\
{\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}} = {{10}^{0.05}}}\\
{\dfrac{{{n_{{\rm{Salt }}}}}}{{{n_{{\rm{Acid }}}}}} = 1.1}\\
{{n_{{\rm{Salt }}}} = 0.0210 \times 1.1}\\
{{n_{{\rm{Salt }}}} = 0.0231{\rm{ mol}}{\rm{. }}}\\
{{m_{{\rm{Salt }}}} = 0.0231{\rm{ mol}}{\rm{. }} \times 144.11\dfrac{g}{{{\rm{ mol}}{\rm{. }}}}}\\
{{m_{{\rm{Salt }}}} \cong 3.3g}
\end{array} $
To achieve the optimal pH of 4.25, 3.3 g of sodium benzoate must be dissolved in the benzoic acid solution.

Note:
The pH scale in chemistry is used to determine the acidity or basicity of an aqueous solution. The pH in acidic solutions is lower than that of basic or alkaline solutions. The pH scale is logarithmic and shows the concentration of hydrogen ions in the solution in inverse order.