Calculate the pH of 0.20 M NH4Cl at 298 K. (K_b of NH3 = 1.8x 10^-5; use pH = 7 - tfrac12(pK_b + log C))

JEE Main levelSingle correct optionDifficulty 2 of 3ia115

Calculate the pH of $0.20 M$ $NH X_{4} Cl$ at 298 K. ( $K_{b}$ of $NH X_{3} = 1.8 \times 1 0^{- 5}$ ; use $pH = 7 - \frac{1}{2} (p K_{b} + lo g C)$ )

NH X_{4} X^{+}

hydrolyses:

[H X^{+}] = K_{w} C / K_{b} = 1 0^{- 14} \times 0.20/ (1.8 \times 1 0^{- 5}) = 1.054 e - 05 M

pH = 4.98