JEE MAIN - Chemistry (2021 - 16th March Morning Shift - No. 15)
Two salts A2X and MX have the same value of solubility product of 4.0 $$\times$$ 10$$-$$12. The ratio of their molar solubilities i.e. $${{S({A_2}X)} \over {S(MX)}}$$ = __________. (Round off to the Nearest Integer)
Answer
50
Explanation
For A2X
$$\matrix{ {{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr {} & {} & {2{S_1}} & {{S_1}} \cr } $$
$${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$$
S1 = 10$$-$$4
for MX
$$\matrix{ {MX} & \to & {{M^ + }} & {{X^ - }} \cr {} & {} & {{S_2}} & {{S_2}} \cr } $$
$${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$$
S2 = 2 $$\times$$ 10$$-$$6
so, $${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$$
$$\matrix{ {{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr {} & {} & {2{S_1}} & {{S_1}} \cr } $$
$${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$$
S1 = 10$$-$$4
for MX
$$\matrix{ {MX} & \to & {{M^ + }} & {{X^ - }} \cr {} & {} & {{S_2}} & {{S_2}} \cr } $$
$${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$$
S2 = 2 $$\times$$ 10$$-$$6
so, $${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$$
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