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Basic Buffer Action

A buffer solu...

Question

A buffer solution is prepared in which the concentration of $N H_{3}$ is $0.30 M$ and the concentration of $N H_{4}^{+}$ is $0.20 M$ . If the equilibrium constant, $K_{b}$ for $N H_{3}$ equals $1.8 \times 10^{- 5}$ , what is the $p H$ of this solution?
$(log 2 / 3 = - 0.176)$

A

9.43

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B

11.72

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C

8.73

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D

9.08

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Solution

The correct option is A $9.43$
$K_{b} = 1.8 \times 10^{- 5}$
$p K_{b} = - l o g K_{b} = - l o g 1.8 \times 10^{- 5} = 4.74$

$p O H = p K_{b} + l o g (\frac{[N H_{4}^{+}]}{[N H_{3}]})$

$p H = 4.74 - l o g (\frac{0.20}{0.30}) = 4.56$

$p H = 14 - 4.56 = 9.436$

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