Ionic EquilibriumMedium

Question

The pH of 0.1 M – N2H4 solution is (For N2H4, Kb1 = 3.6 × 10−6, Kb2 = 6.4 × 10−12, log 2 = 0.3, log 3 = 0.48)

Options

A.3.22
B.2.72
C.10.78
D.11.22

Solution

For POH, 2nd protonation may be neglected

$\left\lbrack OH^{-} \right\rbrack = \sqrt{K_{b_{1}}.C} = \sqrt{3.6 \times 10^{- 6} \times 0.1} = 6 \times 10^{- 4}\text{ M}$

∴ POH = –log(6 × 10–4) = 3.22 ⇒ PH = 10.78

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