Question

A generic metal thiocyanate, M(SCN)2, has a Ksp value of 2.00×10−5. Calculate the molar solubility of the metal thiocyanate in 0.421 M KSCN. Express your answer numerically in units of mM to 4 decimal places.

Answer 1

The molar solubility of the metal thiocyanate is
`1.127* 10^(-4) M`.

Step-by-step explanation:

Concentration of potassium thiocyanate = 0.421 M

Concentration of thiocyanate ion =
`[SCN^-]= 0.421 M`

Concentration of metal ion =
`[M^(2+)]= ?`

The solubility product of metal thiocyanate =
`K_(sp)=2.00* 10^(-5)`

`M(SCN)_2\rightleftharpoons M^(2+)+2SCN^-`

S 2S

At equilbrium

S (2S+0.421)

The expression of solubility product is given by :

`K_(sp)=[M^(2+)]* [SCN^-]^2`

`2.00* 10^(-5)=S* (2S+0.421)^2`

Solving for S:

`S = 1.128* 10^(-4) M`

`[M^(2+)]=(2.00* 10^(-5))/((0.421 M)^2)=1.127* 10^(-4) M`

The molar solubility of the metal thiocyanate is
`1.127* 10^(-4) M`.