First slide

Basic Buffer

Question

Calculate the amount of ${({NH}_{4})}_{2} {SO}_{4}$ in grams which must be added to 500 ml of 0.2 M ${NH}_{3}$ to yield a solution of pH = 9, $K_{b} for {NH}_{3} = 2 \times 10^{- 5}$

Difficult

A

32.48 g
B

42.48 g
C

13.20 g
D

62.48 g

Solution

$P^{OH} = - \log K_{b} + \log \frac{[{NH}_{4}^{+}]}{[{NH}_{4} OH]}$
Let ‘a’ millimoles of ${NH}_{4}^{+}$ is added to a solution
millimoles of ${NH}_{4} OH = 500 \times 0.2 = 100$
$∴ [{NH}_{4}^{+}] = [salt] = a$
Given $P^{H} {= 9 ; P}^{OH} = 5$
$∴ 5 = - \log 2 \times 10^{- 5} + \log \frac{a}{100}$
$∴ a = 200 millimoles = 0.2 moles$
Moles of ${({NH}_{4})}_{2} {SO}_{4}$ added = $\frac{a}{2} = 0.1 mol$
$∴ W {(N H_{4})}_{2} S O_{4} = 0.1 \times 132 = 13.2$ g

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