Question

Calculate the pH of a solution made by adding $0.001\text{mol of}NaOH\text{to}100{\text{cm}}^3$ of solution which is $0.50\text{M}$ in $C{H}_{3}COOH\text{and}0.5\text{M in}C{H}_{3}COONa$.

• A. $6.50$
• B. $7.80$
• C. $3.56$
• D. $4.76$

Answer 1

The correct option is D $4.76$

Amount of $C{H}_{3}COOH$ in $100{\text{cm}}^3$ of the solution $=\frac{\left(0.50\text{mol}\right)}{\left(1000{\text{cm}}^3\right)}\left(100{\text{cm}}^3\right)=0.05\text{mol}$

Amount of $C{H}_{3}COONa$ in $100{\text{cm}}^3$ of the solution $=\frac{\left(0.50\text{mol}\right)}{\left(1000{\text{cm}}^3\right)}\left(100{\text{cm}}^3\right)=0.05\text{mol}$

When alkali is added, the acid is converted into salt. Thus, we have
Amount of the acid after the addition of $0.001\text{mol}$ of the alkali = $0.049\text{mol}$
$\text{Amount of the salt}=0.051\text{mol}$
Since ${\text{pH = pK}}_a^{\circ }+\log \left(\frac{\left(\text{salt}\right)}{\text{(acid)}}\right)$
therefore,
$\text{pH}=-\log (1.8\times {10}^{-5})+\log \left(\frac{0.051}{0.049}\right)=4.7447+0.0174\approx 4.76$