JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 1)
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 $$\times$$ 10$$-$$16 and Ka(HCN) = 6.2 $$\times$$ 10$$-$$10]
1.9 $$\times$$ 10$$-$$5
1.6 $$\times$$ 10$$-$$6
2.2 $$\times$$ 10$$-$$16
0.625 $$\times$$ 10$$-$$6
Explanation
Let solubility is x
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$${K_{sp}} \times {1 \over {{K_a}}} = [A{g^ + }][C{N^ - }] \times {{[HCN]} \over {[{H^ + }][C{N^ - }]}}$$
$$2.2 \times {10^{ - 16}} \times {1 \over {6.2 \times {{10}^{ - 10}}}} = {{[S][S]} \over {{{10}^{ - 3}}}}$$
$${S^2} = {{2.2} \over {6.2}} \times {10^{ - 9}}$$
$${S^2} = 3.55 \times {10^{ - 10}}$$
$$S = \sqrt {3.55 \times {{10}^{ - 10}}} $$
$$S = 1.88 \times {10^{ - 5}} = 1.9 \times {10^{ - 5}}$$
$${K_{sp}} \times {1 \over {{K_a}}} = [A{g^ + }][C{N^ - }] \times {{[HCN]} \over {[{H^ + }][C{N^ - }]}}$$
$$2.2 \times {10^{ - 16}} \times {1 \over {6.2 \times {{10}^{ - 10}}}} = {{[S][S]} \over {{{10}^{ - 3}}}}$$
$${S^2} = {{2.2} \over {6.2}} \times {10^{ - 9}}$$
$${S^2} = 3.55 \times {10^{ - 10}}$$
$$S = \sqrt {3.55 \times {{10}^{ - 10}}} $$
$$S = 1.88 \times {10^{ - 5}} = 1.9 \times {10^{ - 5}}$$
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