Question

Calculate the solubility of Ag₂S (a) in pure water and (b) in a solution in which [S^2-] = 0.129 M. Ksp Ag2S= 1.0x10^-49

Solubility in pure water =	M
Solubility in 0.129 M S2- =	M

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Answer 1

a)

Let solubility in pure water = s

then solubility product = 4 x s² = s x (2s)²

1.0 x 10^-49 = 4 x s²

s² = 0.25 x 10^-49

s = 2.5 x 10^-50

s = 1.58 x 10^-25

b)

Let solubility = s

Conc. of S^2- = 0.129 M

Solubility Product = s x ( 2s + 0.129)²= 1.0 x 10^-49

neglecting 2s in comparison to 0.129 M

s x 0.129² = 1.0 x 10^-49

s = 60.0925 x 10^-49

s = 6.00925 x 10^-50