JEE MAIN - Chemistry (2003 - No. 7)
The solubility in water of a sparingly soluble salt AB2 is 1.0 $$\times$$ 10-5 mol L-1. Its solubility product number will be :
4 $$\times$$ 10-10
1 $$\times$$ 10-15
1 $$\times$$ 10-10
4 $$\times$$ 10-15
Explanation
$$A{B_2}\rightleftharpoons\,{A^{ + 2}} + 2{B^ - }$$
$$\left[ A \right] = 1.0 \times {10^{ - 5}},\,\,$$
$$\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right],$$
$${K_{sp}} = {\left[ B \right]^2}\left[ A \right]$$
$$ = {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right]$$
$$ = 4 \times {10^{ - 15}}$$
$$\left[ A \right] = 1.0 \times {10^{ - 5}},\,\,$$
$$\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right],$$
$${K_{sp}} = {\left[ B \right]^2}\left[ A \right]$$
$$ = {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right]$$
$$ = 4 \times {10^{ - 15}}$$
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