SATHEE: Equilibrium - Result Question 91

Equilibrium - Result Question 91

####95. Solution of $0.1 N N H_{4} O H$ and $0.1 N N H_{4} C l$ has $p H$ 9.25. Then find out $p K_{b}$ of $N H_{4} O H$ [2002]

(a) 9.25

(b) 4.75

(d) 8.25

Show Answer

Solution:

(b) $p O H = p K_{b} + \log \frac{[Salt]}{[Base]}$

or $p K_{b} = p O H - \log \frac{[salt]}{[Base]}$

but $p O H + p H = 14$ or $p O H = 14 - p H$

$∴ 14 - p H - \log \frac{[Salt]}{[Base]} = p K_{b}$

$14 - 9.25 - \log \frac{0.1}{0.1} = p K_{b}$

$14 - 9.25 - 0 = p K_{b}$

$p K_{b} = 4.75$

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Equilibrium - Result Question 91

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