JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(50\, mL\) of \(0.2\, M\) ammonia solution is treated with \(25\, mL\) of \(0.2\, M\, HCl\). If \(pK_b\) of ammonia solution is \(4.75\), the \(pH\) of the mixture will be
- A \(3.75\)
- B \(4.75\)
- C \(8.25\)
- D \(9. 25\)
Answer & Solution
Correct Answer
(D) \(9. 25\)
Step-by-step Solution
Detailed explanation
\(N{H_3} + HCl \to N{H_4}Cl\) moles of \(HCl = 0.2\,M \times 25 \times {10^{ - 3}}\,L = 0.005\) moles \(HCl\) (total consumed) moles of \(N{H_3} = 0.2\,M \times 50 \times {10^{ - 3}}\,L = 0.01\) moles \(HCl\) excess \(N{H_3} = 0.01 - 0.005 = 0.005\) moles \(1\) mole ammonia…
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