Equilibrium - Result Question 62
####64. At $25^{\circ} C$, the dissociation constant of a base, $BOH$, is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in $0.01 M$ aqueous solution of the base would be
[2005]
(a) $1.0 \times 10^{-5} mol L^{-1}$
(b) $1.0 \times 10^{-6} mol L^{-1}$
(c) $2.0 \times 10^{-6} mol L^{-1}$
(d) $1.0 \times 10^{-7} mol L^{-1}$
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Solution:
- (d) Given $K_b=1.0 \times 10^{-12}$
$[BOH]=0.01 M$
$ \alpha=\sqrt{K_b / c} $
$ =\sqrt{\frac{1 \times 10^{-12}}{0.01}}=1.0 \times 10^{-5} $
Now $[OH^{-}]=c . \alpha=0.01 \times 10^{-5}$
$ =1 \times 10^{-7} mol L^{-1} $
65
(c) $CH_3 COOH \rightarrow CH_3 COO^{-}+H^{+}$
$ K_a=\frac{[CH_3 COO^{-}][H^{+}]}{[CH_3 COOH]} $
Given that,
$[CH_3 COO^{-}]=[H^{+}]=3.4 \times 10^{-4} M$
$K_a$ for $CH_3 COOH=1.7 \times 10^{-5}$
$CH_3 COOH$ is weak acid, so in it $[CH_3 COOH]$ is equal to initial concentration. Hence
$ \begin{aligned} & 1.7 \times 10^{-5}=\frac{(3.4 \times 10^{-4})(3.4 \times 10^{-4})}{[CH_3 COOH]} \\ & \begin{aligned} {[CH_3 COOH] } & =\frac{3.4 \times 10^{-4} \times 3.4 \times 10^{-4}}{1.7 \times 10^{-5}} \\ & =6.8 \times 10^{-3} M \end{aligned} \end{aligned} $
