Gujarati
6-2.Equilibrium-II (Ionic Equilibrium)
medium

$0.006\,M$ बेन्जोइक अम्ल के विलयन की हाइड्रोजन आयन सान्द्रता होगी $({K_a} = 6 \times {10^{ - 5}})$

A

$0.6 \times {10^{ - 4}}$

B

$6 \times {10^{ - 4}}$

C

$6 \times {10^{ - 5}}$

D

$3.6 \times {10^{ - 4}}$

Solution

(b) प्रारम्भिक सान्द्रता $= 0.006\, M$ $==> $ ${K_a} = 6 \times {10^{ – 5}}$

साम्य अभिक्रिया ,$\mathop {\mathop {{C_6}{H_5}COOH}\limits_C }\limits_{C(1 – a)\,\,\,} $ $ \rightleftharpoons $ $\mathop {\mathop {{C_6}{H_5}CO{O^ – }}\limits_O }\limits_{Ca} + \mathop {\mathop {{H^ + }\,\,\,\,\,\,\,}\limits_{O\,\,{\rm{initial}}} }\limits_{Ca\,\,\,\,\,\,\,\,\,\,\,\,\,\,} $

${K_a} = C{\alpha ^2}$ $(\because 1 – \alpha \approx 1)$ दुर्बल विधुत अपघट्य के लिए

$\alpha = \sqrt {\frac{{{K_a}}}{C}} $; $\alpha = \sqrt {\frac{{6 \times {{10}^{ – 5}}}}{{0.006}}} $; $\alpha = {10^{ – 1}}$

$\because \,[{H^ + }] = C\alpha $$ = 0.006 \times {10^{ – 1}}$$ = 6 \times {10^{ – 4}}\,M$.

Standard 11
Chemistry

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