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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Two salts $A _{2} X$ and $MX$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=...........$
(Round off to the Nearest Integer).
A
$40$
B
$50$
C
$45$
D
$55$
(JEE MAIN-2021)
Solution
For $A _{2} X$
$A _{2} X \rightarrow 2 A ^{+}+ X ^{2-}$
$2 S _{1} \quad S _{1}$
$K _{ sp }=4 S _{1}^{3}=4 \times 10^{-12}$
$S _{1}=10^{-4}$
for $MX$
$MX \rightarrow M ^{+}+ X ^{-}$
$\begin{array}{ll} S _{2} S _{2}\end{array}$
$K _{ sp }= S _{2}^{2}=4 \times 10^{-12}$
$S _{2}=2 \times 10^{-6}$
so $\frac{ S _{ A _{2} x }}{ S _{ MX }}=\frac{10^{-4}}{2 \times 10^{-6}}=50$
Standard 11
Chemistry
