Question

What is the concentration of silver ions in 3.0 g of solid silver sulfide is added to 100.0 ml of a 0.250 M solution of sodium sulfide Ksp=6.0 x 10^-51

answer is 2.3 x 10^-17 , but unsure how to get it

Answer 1

Concentration of silver sulfide is = ( mass/molar mass) x ( 1000 / volume of solution in mL)

= ( 3.0g / 247.8 (g/mol)) x ( 1000 / 100.0 mL)

= 0.121 M

Ag₂S ---> 2Ag⁺ + S^2-

1 mole of Ag₂S produces 2 moles of Ag⁺

Concentration of Ag⁺ = 2[Ag₂S] = 2 x 0.121 M = 0.242 M

Given solubility product is Ksp = 6.0 x 10^-51

Let S be the solubility of Ag2s

[Ag⁺ ] = 2 x S & [ S^2-] = S

Then Ksp = [Ag⁺ ]² [ S^2-]

6.0 x 10^-51 = (2S)² S

                = 4S³

S = 1.14x10^-17 M