The expression to compute pH of ${NH}_{4} Cl$ solution is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$pH = \frac{1}{2} p K_{w}^{\circ} - \frac{1}{2} p K_{b}^{\circ} ({NH}_{4} OH) - \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} p K_{b}^{\circ} ({NH}_{4} OH) - \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} p K_{b}^{\circ} ({NH}_{4} OH) + \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{b}^{\circ} ({NH}_{4} OH) - \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} \log ([salt] / M)$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} {NH}_{4}^{+} + H_{2} O ⇌ {NH}_{4} OH + H^{+} \\ K_{h} = \frac{K_{w}}{K_{b}} = \frac{[{NH}_{4} OH] [H^{+}]}{[{NH}_{4}^{+}]} ≃ \frac{{[H^{+}]}^{2}}{{[{NH}_{4}^{+}]}_{0}} \end{array}$

or ${[H^{+}]}^{2} = \frac{K_{w}}{K_{b}} {[{NH}_{4}]}_{0}$

$2 \log ([H^{+}] / M) = \log K_{w}^{\circ} - \log K_{b}^{\circ} + \log ([{NH}_{4}^{+}] / M)$ or $- \log ([H^{+}] / M) = - \frac{1}{2} \log K_{W}^{\circ} + \frac{1}{2} \log K_{b}^{\circ} - \frac{1}{2} \log ([{NH}_{4}^{+}] / M)$