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 (integer)
50
Solution
For A<sub>2</sub>X<br><br>$$\matrix{
{{A_2}X} & \to & {2{A^ + }} & {{X^{2 - }}} \cr
{} & {} & {2{S_1}} & {{S_1}} \cr
} $$<br><br>${K_{sp}} = 4S_1^3 = 4 \times {10^{ - 12}}$<br><br>S<sub>1</sub> = 10<sup>$-$4</sup><br><br>for MX<br><br>$$\matrix{
{MX} & \to & {{M^ + }} & {{X^ - }} \cr
{} & {} & {{S_2}} & {{S_2}} \cr
} $$<br><br>${K_{sp}} = S_2^2 = 4 \times {10^{ - 12}}$<br><br>S<sub>2</sub> = 2 $\times$ 10<sup>$-$6</sup><br><br>so, $${{{S_{{A_2}X}}} \over {{S_{MX}}}} = {{{{10}^{ - 4}}} \over {2 \times {{10}^{ - 6}}}} = 50$$
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
This question is part of PrepWiser's free JEE Main question bank. 67 more solved questions on Equilibrium are available — start with the harder ones if your accuracy is >70%.