Equilibrium - Result Question 44
####46. The ionization constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 K$. Hydrolysis constant of ammonium chloride is:
(a) $6.50 \times 10^{-12}$
(b) $5.65 \times 10^{-13}$
(c) $5.65 \times 10^{-12}$
(d) $5.65 \times 10^{-10}$
[2009]
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Solution:
- (d) Ammonium chloride is a salt of weak base and strong acid. In this case, hydrolysis constant, $K_h$ can be calculated as
$ K_h=\frac{K_w}{K_b}=\frac{1 \times 10^{-14}}{1.77 \times 10^{-5}}=5.65 \times 10^{-10} $
