Answer: The molar solubility of MX is [tex]2.54\times 10^{-36}M[/tex]
Explanation:
We are given:
[tex]K_{sp}(MX)=1.27\times 10^{-36}[/tex]
The chemical equation for the ionization of MX follows:
[tex]MX\rightleftharpoons M^+(aq.)+X^-(aq.)[/tex]
S S
The expression of [tex]K_{sp}[/tex] for above equation is:
[tex]K_{sp}=S\times S[/tex] ......(1)
The chemical equation for the ionization of [tex]M_2SO_4[/tex] follows:
[tex]M_2SO_4\rightleftharpoons 2M^+(aq.)+SO_4^{2-}(aq.)[/tex]
0.25M 0.5M 0.25M
Total concentration of cation from both the equation is:
[tex][M^+]=0.5+S[/tex]
As, [tex]K_{sp}(MX)<<1[/tex], so S is also very very less than 1 and can be easily neglected.
So, [tex][M^+]=0.5M[/tex]
Putting values in equation 1, we get:
[tex]1.27\times 10^{-36}=0.5\times S\\\\S=2.54\times 10^{-36}M[/tex]
Hence, the molar solubility of MX is [tex]2.54\times 10^{-36}M[/tex]
