ExamBro
ExamBro
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 \(log\,2 = 0.301\) ]

  1. A \(2.65\)
  2. B \(5.35\)
  3. C \(4.35\)
  4. D \(4.65\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5.35\)

Step-by-step Solution

Detailed explanation

For the salt of strong acid and weak base \({H^ + } = \sqrt {\frac{{{K_w} \times C}}{{{K_b}}}} \) \([{H^ + }] = \sqrt {\frac{{{{10}^{ - 14}} \times 2 \times {{10}^{ - 2}}}}{{{{10}^{ - 5}}}}} \) \( - \log [{H^ + }] = 6 - \frac{1}{2}\log \,20\) \(\therefore \,pH = 5.35\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app