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* 10^(-36)M`

Step-by-step explanation:

We are given:

`K_(sp)(MX)=1.27* 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* 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)<<1`, 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* 10^(-36)=0.5* S\\\\S=2.54* 10^(-36)M`

Hence, the molar solubility of MX is
`2.54* 10^(-36)M`