JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The \(pH\) of a \(0.02\,M\,\,NH_4Cl\) solution will be [given \(K_b\,(NH_4OH) = 10^{-5}\) and \(log\,2 = 0.301\) ]
- A \(2.65\)
- B \(5.35\)
- C \(4.35\)
- D \(4.65\)
Answer & Solution
Correct Answer
(B) \(5.35\)
Step-by-step Solution
Detailed explanation
For the salt of strong acid and weak base \({H^ + } = \sqrt {\frac{{{K_w} \times C}}{{{K_b}}}} \) \([{H^ + }] = \sqrt {\frac{{{{10}^{ - 14}} \times 2 \times {{10}^{ - 2}}}}{{{{10}^{ - 5}}}}} \) \( - \log [{H^ + }] = 6 - \frac{1}{2}\log \,20\) \(\therefore \,pH = 5.35\)
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