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6-2.Equilibrium-II (Ionic Equilibrium)
medium
$0.01\,M$ $HCN$ तथा $0.02\,M$ $NaCN$ का एक विलयन है जिसमें $[{H^ + }]$ का सान्द्रण होगा$\left( {HCN} \right.$ के लिये ${K_a}$ $ = 6.2 \times {10^{ - 10}})$
A
$3.1 \times {10^{10}}$
B
$6.2 \times {10^5}$
C
$6.2 \times {10^{ - 10}}$
D
$3.1 \times {10^{ - 10}}$
Solution
(d) ${K_a} = \frac{{[{H^ + }]\,\,[C{N^ – }]}}{{[HC{N^ – }]}}$
$6.2 \times {10^{ – 10}} = \frac{{[{H^ + }]\,\,[0.02]}}{{[0.01]}}$
$[{H^ + }] = \frac{{6.2 \times {{10}^{ – 10}} \times 0.01}}{{0.02}} = 3.1 \times {10^{ – 10}}$
Standard 11
Chemistry
