Equilibrium - Result Question 79
####83. A buffer solution is prepared in which the concentration of $NH_3$ 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? $(\log 2.7=0.433)$.
(a) 9.08
(b) 9.43
(c) 11.72
(d) 8.73
[2011]
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Solution:
- (b) Given : $[NH_3]=0.3 M,[NH_4^{+}]=0.2 M$, $K_b=1.8 \times 10^{-5}$.
$pOH=p K_b+\log \frac{[\text{ salt }]}{[\text{ base }]}$
$ [p K_b=-\log K_b ; p K_b=-\log 1.8 \times 10^{-5}] $
$\therefore p K_b=4.74$
$ \begin{aligned} pOH & =4.74+\log \frac{0.2}{0.3} \\ & =4.74+0.3010-0.4771=4.56 \\ pH & =14-4.56=9.436 \end{aligned} $
