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A buffer solution is prepared in which the conc. ${NH}_{3}$ is 0.30 M and the concentration of ${NH}_{4}^{+}$ is 0.20 M. If the equilibrium constant, $K_{b} for {NH}_{3}$ equals $1.8 \times 10^{- 5}$ , what is the pH of this solution

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9.08

11.72

9.44

8.73

answer is C.

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Detailed Solution

$pOH = {pK}_{b} + \log \frac{[salt]}{[base]} = 4.74 + \log \frac{0.20}{0.30}$

= 4.74 + (0.301- 0.477)

= 4.74 - 0.176 = 4.56

pH = 14 - 4.56 = 9.44

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