First slide

Ionic equilibrium

Question

${pK}_{b} of aq. {NH}_{3} is 4 .74, hence pH of 0 .01 M {NH}_{3} solution is$

Moderate

A

3.37
B

10.63
C

2.00
D

12.00

Solution

$[{OH}^{-}] = \sqrt{K_{b} C}$
$∴ pOH = \frac{1}{2} [{pK}_{b} - logC] = 3 .37 ∴ pH = 14 - pOH = 10 .63$

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