MCQ
If $20\, ml$ of $0.1\, M \,NaOH$ is added to $30\, ml$ of $0.2\, M \,CH_3COOH (pK_a = 4.74)$. Find $pH$ of resulting solution :-
- A$4.44$
- B$5.48$
- C$4.74$
- ✓$6.44$
✓
Answer
Correct option: D.
$6.44$
d
$\mathop {\mathop {NaOH(aq)}\limits_{20 \times 0.1} }\limits_{2\,m\,mole} + \mathop {\mathop {C{H_3}COOH(aq)}\limits_{30 \times 0.2} }\limits_{6\,m\,mole} \to $ $\mathop {\mathop {C{H_3}COONa(aq)}\limits_0 }\limits_{2\,m\,mole} + \mathop {\mathop {{H_2}O(l)}\limits_0 }\limits_{2\,m\,mole} $
$\mathop {\mathop {NaOH(aq)}\limits_{20 \times 0.1} }\limits_{2\,m\,mole} + \mathop {\mathop {C{H_3}COOH(aq)}\limits_{30 \times 0.2} }\limits_{6\,m\,mole} \to $ $\mathop {\mathop {C{H_3}COONa(aq)}\limits_0 }\limits_{2\,m\,mole} + \mathop {\mathop {{H_2}O(l)}\limits_0 }\limits_{2\,m\,mole} $
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \frac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{COONa}}} \right]}}{{\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right]}}$
$ = 4.74 + \log \frac{2}{4}$
$ = 4.74 - 0.3010$
$pH = 4.44$
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