Q. The K_b of NH₄OH is 1.77 x 10^{- 5} at 298 K. The value of hydrolysis constant of ammonium chloride is:

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A

$5 \text{.} 6 5 \times 1 0^{- 1 2}$

B

$5 \text{.} 6 5 \times 1 0^{- 1 0}$

C

$6 \text{.} 5 0 \times 1 0^{- 1 2}$

D

$5 \text{.} 6 5 \times 1 0^{- 1 3}$

Solution:

$\left(\text{NH}\right)_{4}\left(\text{Cl}\right)_{\left(\text{aq}\right)} & + & \left(\text{H}\right)_{2}\text{O} & \rightleftharpoons & \left(\text{NH}\right)_{4}\text{OH} & + & \text{HCl} \\ \text{NH}_{4 \left(\text{aq}\right)}^{+} & + & \left(\text{H}\right)_{2}\text{O} & \rightleftharpoons & \left(\text{NH}\right)_{4}\text{OH} & + & \text{H}_{\left(\text{aq}\right)}^{+}$
$\text{K}_{\text{h}} = \frac{\text{K}_{\text{w}}}{\text{K}_{\text{b}}} = \frac{1 0^{- 1 4}}{1 \text{.} 7 7 \times 1 0^{- 5}} = 5 \text{.} 6 5 \times 1 0^{- 1 0}$

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