SATHEE: Equilibrium - Result Question 79

Equilibrium - Result Question 79

####83. 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? $(\log 2.7 = 0.433)$ .

(a) 9.08

(b) 9.43

(c) 11.72

(d) 8.73

[2011]

Show Answer

Solution:

(b) Given : $[N H_{3}] = 0.3 M, [N H_{4}^{+}] = 0.2 M$ , $K_{b} = 1.8 \times 10^{- 5}$ .

$p O H = p K_{b} + \log \frac{[salt]}{[base]}$

$[p K_{b} = - \log K_{b}; p K_{b} = - \log 1.8 \times 10^{- 5}]$

$∴ p K_{b} = 4.74$

$\begin{aligned} p O H & = 4.74 + \log \frac{0.2}{0.3} \\ = 4.74 + 0.3010 - 0.4771 = 4.56 \\ p H & = 14 - 4.56 = 9.436 \end{aligned}$

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Equilibrium - Result Question 79

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