Question

The mass of manganese(II) sulfide that is dissolved in 150 mL of a saturated solution is _____ grams.

Ksp= 5.1 × 10^-15

Answer 1

The Manganese (II) sulfide dissociates in water according to following equation

MnS (s) Mn²⁺ (aq) + S^2- (aq)

The expression of Ksp is as

Ksp = [ Mn²⁺ ][S^2- ]

Suppose at equilibrium, [ Mn²⁺ ] = [S^2- ] = s

Ksp = 5.1 x 10^-15

Ksp = (s)(s)

s = (Ksp)^1/2 = (5.1 x 10^-15)^1/2

s = 7.14 x 10^-8

[Mn²⁺] = [S^2-] = 7.14 x 10^-8

volume of solution = 150 mL = 0.150 L

No. of moles of  [Mn²⁺] = molarity x volume (L) = 7.14 x 10^-8 x 0.150 L = 1.07 x 10^-8 moles

No. of moles of  [Mn²⁺] = No. of moles of [S^2-] = No. of moles of [MnS] = 1.07 x 10^-8 moles

molar mass of MnS = 87.00 g/mole

Mass of MnS = (No. of moles of MnS) x molar mass of MnS = 1.07 x 10^-8 moles x 87.00 g/mole

= 9.32 x 10^-7 g