What us the solubility in mol/L of manganese(II) sulfide (MnS) given that its Ksp value is 1.7 x10^-13?

Question

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Answer 1

MnS(s) <---> Mn²⁺(aq) + S^2-(aq)

ksp=[Mn²⁺][S^2-]

Let x=molarity of Mn²⁺,

Then molarity of S^2- is also x due to 1:1 molar ratio between Mn²⁺ and S^2-.

Substitute x for [Mn²⁺] and [S^2-]

ksp=x*x or ksp=x²

1.7*10^-13=x²

Find square root of x

4.123*10^-7 M=x

This is the molar solubility of Mn^2+.