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Introduction and Degree of Hydrolysis

Calculate the...

Question

Calculate the weight of $(N H_{4})_{2} S O_{4}$ which must be added to 500 mL of 0.2 M $N H_{3}$ to yield a solution of pH = 9.35, $K_{b}$ for $N H_{3} = 1.78 \times 10^{- 5}$ (in gm)(Write answer nearest integer)

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Solution

$p O H = - l o g K_{b} + l o g \frac{[S a l t]}{[B a s e]}$

$p O H = - l o g K_{b} + l o g \frac{[N H_{4}^{\oplus}]}{[N H_{4} O H]}$

$∵ [N H_{4}^{\oplus}]$ is obtained from salt $(N H_{4})_{2} S O_{4}$

$∵ p H = 9.35$ , therefore, $p O H = 14 - 9.35 = 4.65$

$∴$ mmol of $N H_{4} O H$ in solution $= 0.2 \times 500 = 100$

Let a millimoles of $N H_{4}^{\oplus}$ are added in solution.

$[N H_{4}^{\oplus}] = \frac{a}{500}, [N H_{4} O H] = \frac{100}{500}$

$4.65 = - l o g (1.78 \times 10^{- 5}) + l o g \frac{a / 500}{100 / 500}$
$4.65 = 4.7496 + l o g \frac{a}{100} ∴ a = 79.51$

$∴$ mmol of $(N H_{4})_{2} S O_{4}$ added $= \frac{a}{2} = \frac{79.51}{2} = 39.755$

$∴ \frac{W}{132} \times 1000 = 39.755 ∴ W_{(N H_{4})_{2} S O_{4}} = 5.248 g$

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