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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 pH of this solution? ( $log 2 = 0.3010$ , $log 3 = 0.4771$ )

9.08

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9.43

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11.72

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8.73

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Solution

The correct option is B $9.43$
Given,
$[N H_{3}] = 0.3 M$ , $[N H_{4}^{+}] = 0.2 M$
$K_{b} = 1.8 \times 10^{- 5}$
$p O H = p k_{b} + l o g \frac{[salt]}{[base]}$

Substituting values in the above expression, we get

$p O H = 4.74 + l o g \frac{0.2}{0.3}$ $= 4.74 + 0.3010 - 0.4771 = 4.56$
$p H = 14 - 4.56 = 9.436$

Hence, the pH of the solution is $9.43$ .

Suggest Corrections

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