JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(20\, mL\) of \(0.1\, M\, H_2SO_4\) is added to \(30\, mL\) of \(0.2\, M\, NH_4OH\) solution. The \(pH\) of the resultant mixture is [\(pk_b\) of \(NH_4OH = 4.7\)]
- A \(5.2\)
- B \(9\)
- C \(5\)
- D \(9.4\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(\mathop {\mathop {{H_2}S{O_4}}\limits_{2\,m.m} }\limits_ - + \mathop {\mathop {2N{H_4}OH}\limits_{6\,m.m} }\limits_{2\,m.m} \to \mathop {{{(N{H_4})}_2}S{O_4}}\limits_{2\,m.m} + 2{H_2}O\) \(pOH = 4.7 + \log \frac{4}{2} = 5\) \(pH = 14 - 5 = 9\)
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