Question

To a 50 mL of $0.05{\text{M}}$ formic acid how much volume of $0.10{\text{M}}$sodium formate must be added to get a buffer solution of pH =$4.0$? (${\text{p}}{{\text{K}}_{\text{a}}}$of the acid is $3.8$)
A.50 mL
B.4 mL
C.$39.6$mL
D.100 mL

Answer 1

To solve this question knowledge on the Henderson-Hasselbalch equation is required. This equation is used to determine the pH of a buffer solution. The value of the acid dissociation constant, ${{\text{K}}_{\text{a}}}$ is assumed and the pH is calculated for the concentration of the acid and its conjugate base or base and its conjugate acid.
Formula used:
${\text{molarity = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution}}}}$
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + lo}}{{\text{g}}_{{\text{10}}}}\left[ {\dfrac{{{\text{conjugate base}}}}{{{\text{acid}}}}} \right]$

Complete step by step answer:
In the given question, the acid is formic acid and the conjugate base is sodium formate. As we know that:
${\text{moles of solute = vol of solution}} \times {\text{molarity}}$
Let V mL of $0.10{\text{M}}$sodium formate be added to 50 mL of $0.05{\text{M}}$formic acid:
The molarity of the solution for sodium formate = $\dfrac{{{\text{V}} \times 0.1}}{{\left( {{\text{V}} + 50} \right)}} = \dfrac{{0.1{\text{V}}}}{{{\text{V}} + 50}}$
And the molarity for formic acid = $\dfrac{{50 \times 0.5}}{{\left( {{\text{V}} + 50} \right)}} = \dfrac{{25}}{{{\text{V}} + 50}}$
Putting the values of the concentrations in the Henderson-Hasselbalch equation we get,
${\text{ }} \Rightarrow {\text{4 = 3}}{\text{.8 + lo}}{{\text{g}}_{{\text{10}}}}\left[ {\dfrac{{\dfrac{{0.1{\text{V}}}}{{\left( {{\text{V}} + 50} \right)}}}}{{\dfrac{{25}}{{\left( {{\text{V}} + 50} \right)}}}}} \right]$
Where, the pH of the buffer is 4 and ${\text{p}}{{\text{K}}_{\text{a}}}$of the acid is $3.8$
Solving the above equation we get,
${\text{ lo}}{{\text{g}}_{{\text{10}}}}\left[ {\dfrac{{0.1{\text{V}}}}{{25}}} \right]{\text{ = 4 - 3}}{\text{.8 }}$
$\Rightarrow {\log _{10}}{\text{V}} - 2.397 = 0.2$
Solving for volume, we get:
$\Rightarrow {\text{V}} = 49.88{\text{ mL }} \sim 50{\text{ mL}}$

So, the correct answer is option A, 50 mL.

Note:
A simple buffer system consists of a weak acid and the salt of the conjugate base of that acid.
There are two assumptions on which the Henderson-Hasselbalch equation is based upon. Firstly, the acid is a monobasic acid and dissociates according to the equation:
${\text{HA}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + }}{{\text{A}}^{\text{ - }}}$
And secondly, the self-ionization of water is ignored and not included in the equation.
As this cannot be ignored in case of the strong acids, so the Henderson-Hasselbalch equation is not applicable to strong acids.