Question

A ${{\text{K}}_{\text{a}}}$ for acetic acid is $1.7 \times {10^{ - 5}}$ at ${25^0}{\text{C}}$. The pH of a mixture of $25{\text{ mL}}$ $0.02{\text{N}}$ acetic acid and $2.5{\text{mL}}$ $0.1{\text{N}}$ ${\text{NaOH}}$ (neglecting volume change) will be ($\log 1.7 = 0.23$)
A.$2.2$
B.$4.8$
C.$7.5$
D.$1.0$

Answer 1

The ${{\text{K}}_{\text{a}}}$ of the acid is the acid dissociation constant that determines the ability of the acid to release the proton. The higher the value of ${{\text{K}}_{\text{a}}}$, the higher will be the degree of dissociation.
Formula Used:
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log }}\dfrac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$, where ${\text{p}}{{\text{K}}_{\text{a}}}$= $- {\log _{10}}\left[ {{{\text{K}}_{\text{a}}}} \right]$

Complete step by step answer:
When acetic acid reacts with sodium hydroxide, it leads to the formation of sodium acetate, which is a basic salt of acetic acid and sodium hydroxide, and water.
According to the Henderson equation, the relation between the pH of the solution and the ${\text{p}}{{\text{K}}_{\text{a}}}$of the solution I given by the following relation:
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log }}\dfrac{{\left[ {{\text{salt}}} \right]}}{{\left[ {{\text{acid}}} \right]}}$, where ${\text{p}}{{\text{K}}_{\text{a}}}$= $- {\log _{10}}\left[ {{{\text{K}}_{\text{a}}}} \right]$
Hence, ${\text{p}}{{\text{K}}_{\text{a}}}$= $- {\log _{10}}\left[ {{\text{1}}{\text{.7}}} \right] - \left( { - 5{{\log }_{10}}10} \right)$=$5 - 0.23 = 4.8$.
The total volume of the solution, after adding the base to the acid is,
$25 + 2.5 = 27.5{\text{mL}}$= $0.0275{\text{L}}$
The concentration of the salt = $\dfrac{{0.0025 \times 0.1}}{{0.0275}} = 0.009{\text{mol }}{{\text{L}}^{ - 1}}$, and
The concentration of the salt = $\dfrac{{\left( {0.025 \times 0.2} \right) - \left( {0.0025 \times 0.1} \right)}}{{0.0275}} = 0.009{\text{mol }}{{\text{L}}^{ - 1}}$
Now putting the values of the concentrations of the salt formed and the acid, we get,
${\text{pH = 4}}{\text{.8 + log }}\dfrac{{\left[ {0.09} \right]}}{{\left[ {0.09} \right]}}$
$\Rightarrow {\text{pH = 4}}{\text{.8 + log }}1 = 4.8 + 0 = 4.8$.

Hence, the correct answer is option B.

Note:
The salt of a weak acid and a strong base is always a basic salt because the salt dissociates in water to liberate hydroxyl anions which the weak acid stays mostly undissociated. Similarly the salt of a weak base and a strong acid is always an acidic salt that dissociates in water to liberate hydronium ions while the weak base stays undissociated mostly.