What is the pH of a 0.20 M solution of NH4Cl at 298 K? K_b(NH3) = 1.8x10^-5. Use pH = 7 - tfrac12(pK_b + log c).

JEE Main levelSingle correct optionDifficulty 2 of 3ia006

What is the pH of a 0.20 M solution of $NH X_{4} Cl$ at 298 K? $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Use $pH = 7 - \frac{1}{2} (p K_{b} + lo g c)$ .

NH X_{4} X^{+}

hydrolyses with

K_{h} = K_{w} / K_{b} = 5.56 e - 10

and

[H X^{+}] = K_{h} c

. Hence

pH = 7 - \frac{1}{2} (p K_{b} + lo g c) = 7 - \frac{1}{2} (4.745 - 0.699) = 4.98