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pH of a Solution

Calculate the...

Question

Calculate the amount of $(N H_{4})_{2} S O_{4}$ in g which must be added to $500$ mL of $0.2$ M $N H_{3}$ to yield a solution of $p H = 9.35$ $K_{b}$ for $N H_{3} = 4.7$

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Solution

Molarity of $N H_{3}$ solution = $\frac{moles of N H_{3}}{volume of solution}$
$moles of N H_{3} = 0.2 \times 0.5 = 0.1$
$p O H = 14 - p H$
$p O H = 14 - 9.35 = 4.65$
$p K_{b} = - l o g [K_{b}] = - l o g [1.78 \times 10^{- 5}] = 4.74$
$p O H = p K_{b} - l o g \frac{[b a s e]}{[s a l t]}$
$l o g \frac{[b a s e]}{[s a l t]} = 4.74 - 4.65 = 0.09$
$\frac{[b a s e]}{[s a l t]} = a n t i l o g [0.09] = 1.23$
[base] = moles of $N H_{3} = 0.1$
$[s a l t] = \frac{0.1}{1.23} = 0.08$
moles of ${(N H_{4})}^{+} = 0.08$
${(N H_{4})}_{2} S O_{4} ⇌ {2 N H_{4}}^{+} + {(S O_{4})}^{- 2}$
$moles of {(N H_{4})}_{2} S O_{4} = \frac{moles of {N H_{4}}^{+}}{2} = 0.04$
$amount of {(N H_{4})}_{2} S O_{4} = 0.04 \times 132 = 5.28 g$

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