Equilibrium - Result Question 91
####95. Solution of $0.1 N NH_4 OH$ and $0.1 NNH_4 Cl$ has $pH$ 9.25. Then find out $p K_b$ of $NH_4 OH$ [2002]
(a) 9.25
(b) 4.75
(c) 3.75
(d) 8.25
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Solution:
(b) $pOH=p K_b+\log \frac{[\text{ Salt }]}{[\text{ Base }]}$
or $p K_b=pOH-\log \frac{\text{ [salt] }}{\text{ [Base] }}$
but $pOH+pH=14$ or $pOH=14-pH$
$\therefore 14-pH-\log \frac{[\text{ Salt }]}{[\text{ Base }]}=p K_b$
$14-9.25-\log \frac{0.1}{0.1}=p K_b$
$14-9.25-0=p K_b$
$p K_b=4.75$
