100 mL of 0.20 M NH3 is mixed with 50 mL of 0.15 M HCl. For NH3, K_b = 1.8 x 10^-5 (pK_b = 4.745). Find the pH of the mixture (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ia101

$100 mL$ of $0.20 M$ $NH X_{3}$ is mixed with $50 mL$ of $0.15 M$ $HCl$ . For $NH X_{3}$ , $K_{b} = 1.8 \times 1 0^{- 5}$ ( $p K_{b} = 4.745$ ). Find the pH of the mixture (two decimals).

Numerical answer type, tolerance 0.01.

NH X_{3}

: 20 mmol;

HCl

: 7.5 mmol. After reaction

NH X_{3}

= 12.5 mmol and

NH X_{4} X^{+}

= 7.5 mmol.

pOH = p K_{b} + lo g ([NH X_{4} X^{+}] / [NH X_{3}]) = 4.745 + lo g (7.5/12.5) = 4.523

, so pH = 9.48.

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