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Basic Buffer Action

A buffer of ...

Question

A buffer of $p H = 9.26$ is made by dissolving $x$ moles of ammonium sulphate and 0.1 mole of ammonia into 100 mL solution. If $p K_{b}$ of ammonia is 4.74, calculate value of $x$ .

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Solution

$p H = p K_{b} + log \frac{[s a l t]}{[b a s e]}$
Given, $p H = 9.26, p K_{b} = 4.76$
$[s a l t] = x m o l e / L$
$[b a s e] = (0.1 \times 1000) / 100 = 1 M o l / L$
Now, $9.26 = 4.74 + log \frac{x}{1}$
$\Rightarrow 4.52 = log x$
$\Rightarrow x = 10^{- 4.52} = 3.02 \times 10^{- 5}$ .

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