Question

A mysterious compoud with the chemical formula MX (Ksp 1.27x10-36) is present in two different solutions. What is its molar solubility in 0.25 M M2SO4?

Answer 1

Answer: The molar solubility of MX is $2.54\times 10^{-36}M$

Explanation:

We are given:

$K_{sp}(MX)=1.27\times 10^{-36}$

The chemical equation for the ionization of MX follows:

$MX\rightleftharpoons M^+(aq.)+X^-(aq.)$

                  S         S

The expression of $K_{sp}$ for above equation is:

$K_{sp}=S\times S$      ......(1)

The chemical equation for the ionization of $M_2SO_4$ follows:

$M_2SO_4\rightleftharpoons 2M^+(aq.)+SO_4^{2-}(aq.)$

 0.25M           0.5M      0.25M

Total concentration of cation from both the equation is:

$[M^+]=0.5+S$

As, $K_{sp}(MX)$, so S is also very very less than 1 and can be easily neglected.

So, $[M^+]=0.5M$

Putting values in equation 1, we get:

$1.27\times 10^{-36}=0.5\times S\\\\S=2.54\times 10^{-36}M$

Hence, the molar solubility of MX is $2.54\times 10^{-36}M$