Question

The Ksp for silver sulfate (Ag2SO4) is 1.2 × 10-5. Calculate the molar solubility of silver sulfate in a solution of (4.5x10^-1) M K2SO4 Please clearly show your work and explain steps

Answer 1

Given that Ksp of (Ag₂SO₄) = 1.2 × 10^-5

[K₂SO₄ ] = [SO₄2-] = 4.5x10^-1 M = 0.45 M

Ag2SO4 -------------> 2Ag⁺ + SO₄^2-

                                  2s             s                  s = molar solubility of silver sulfate

Ksp = [ 2Ag⁺]² [ SO₄^2-]

  1.2 × 10^-5 = [2s]² [ s + 0.45 ]

1.2 × 10^-5 =   [ 4s³ + 1.8 s² ]

Since s is very small, neglect s³ term

Hence,

1.2 × 10^-5 =    1.8 s²

       s = 0.258 x 10^-2 M

Therefore,

molar solubility of silver sulfate = 0.258 x 10^-2 M