The basic ionisation constant for hydrazine, N 2 H 4 is 9.6 × 1 0 − 7 . What would be the percent hydrolysis of 0.1 N 2 H 5 Cl ?

Question

The basic ionisation constant for hydrazine, N 2 ​ H 4 ​ is 9.6 × 1 0 − 7 . What would be the percent hydrolysis of 0.1 N 2 ​ H 5 ​ Cl ?

TestHub · Accepted Answer

The salt N₂H₅Cl undergoes cationic hydrolysis:

N₂H₅⁺ + H₂O ⇌ N₂H₄ + H₃O⁺

Hydrolysis constant, $K_{h} = \frac{K _{w}}{K _{b}} = \frac{1 0 ^{- 14}}{9.6 \times 1 0 ^{- 7}} = 1.04 \times 1 0^{- 8}$

$K_{h} = \frac{h ^{2}}{c}$ , where $h$ is the degree of hydrolysis and $c$ is the concentration.

$h = \frac{K _{h}}{c} = \frac{1.04 \times 1 0 ^{- 8}}{0.1} = 1.04 \times 1 0^{- 7} = 3.22 \times 1 0^{- 4}$

Percent hydrolysis = $h \times 100 = 3.22 \times 1 0^{- 4} \times 100 = 0.0322%$

Therefore, the percent hydrolysis of 0.1 M N₂H₅Cl is approximately 0.032%.

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