Question

Find the change in the pH when ${\text{0}} \cdot {\text{01 Mol}}$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$ is added to one litre of ${\text{0}} \cdot {\text{01 M}}$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$. (${\text{p}}{{\text{K}}_{\text{a}}} = 4.74$)
A) ${\text{3}} \cdot {\text{27}}$
B) ${\text{4}} \cdot {\text{74}}$
C) ${\text{1}} \cdot {\text{37}}$
D) ${\text{2}} \cdot {\text{74}}$

Answer 1

The given problem can be solved by using the Henderson-Hasselbalch equation given below: ${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{{\text{con}}{{\text{c}}^n}{\text{.}}\;{\text{of}}\;{\text{salt}}}}{{{\text{con}}{{\text{c}}^n}{\text{.}}\;{\text{of}}\;{\text{acid}}}}$. One can put the correct values and can calculate the change in pH to make the correct choice of answer.

Complete step by step answer: 1) First of all for the determination of the change in pH, we need to first calculate the concentration of salt $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}} \right)$ from the given moles and volume as below,
Given data,
Moles of the salt, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa = 0}} \cdot {\text{01 Mol}}$
${\text{Volume of salt }} = 1L$
2) Now let us see the formula for the calculation of concentration as below,
${\text{Concentration of C}}{{\text{H}}_{\text{3}}}{\text{COONa = }}\dfrac{{{\text{moles}}\;{\text{of}}\;{\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}}}{{{\text{volume}}}}$
Now let us put the known values in the above formula we get,
${\text{Concentration of C}}{{\text{H}}_{\text{3}}}{\text{COONa}} = \dfrac{{{\text{0}}{\text{.01}}\;{\text{mol}}}}{{{\text{1}}\;{\text{L}}}}$
By doing the calculation part we get,
${\text{Concentration of C}}{{\text{H}}_{\text{3}}}{\text{COONa}} = 0 \cdot 01mol/L{\text{ or 0}} \cdot {\text{01 M}}$
3) Now, let us substitute the values in the Henderson-Hasselbalch equation we get,
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{{\text{conc}}{\text{.}}\;{\text{of}}\;{\text{salt}}}}{{{\text{conc}}{\text{.}}\;{\text{of}}\;{\text{acid}}}}$
By putting the known values in the above equation we get,
$pH = 4.74 + \log \dfrac{{{\text{0}}{\text{.01}}\;{\text{M}}}}{{{\text{0}}{\text{.01}}\;{\text{M}}}}$
Now by doing the calculation we get,
$pH = 4.74 + \log 1$
As the value of ${\text{log1}}$ is zero we get the above equation as below,
$pH = 4 \cdot 74$
4) Therefore, the change in the pH when ${\text{0}} \cdot {\text{01 Mol}}$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$ is added to one liter of ${\text{0}} \cdot {\text{01 M}}$ ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ is ${\text{4}} \cdot {\text{74}}$ which shows option C as the correct choice of answer.

Note:
From the above calculated values, it can be seen that the pH and ${\text{p}}{{\text{K}}_{\text{a}}}$ values are the same. When the concentrations of acid and the conjugate base or salt are the same, i.e. when the acid is ${\text{50\% }}$ dissociated, the pH will be equal to the ${\text{p}}{{\text{K}}_{\text{a}}}$ of acid. From the above calculation, the pH value is coming out to be the same as that of ${\text{p}}{{\text{K}}_{\text{a}}}$ value which means the concentration is the same and there is no change in pH.