20.0 mL of 0.100 M NH3 (K_b = 1.8x 10^-5) is titrated with 0.100 M HCl. The pH at the equivalence point is

JEE Main levelSingle correct optionDifficulty 2 of 3ia134

$20.0 mL$ of $0.100 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) is titrated with $0.100 M$ HCl. The pH at the equivalence point is

At equivalence the solution is

0.0500 M

NH X_{4} Cl

(

2.0 mmol

40 mL

pH = 7 - \frac{1}{2} (p K_{b} + lo g C) = 7 - \frac{1}{2} (4.745 - 1.301) = 5.28