The molar solubility of silver sulfate in a 0.142 M ammonium sulfate solution is _______ M

Expert Solution

The Ksp of silver sulfate is 1.4×10^-5

Ag2SO4 = 2Ag⁺(aq) + SO₄^2-(aq)

0.142 M ammonium sulfate has 0.142 M sulfate

Ksp = [Ag+]²[SO₄^2-]

1.4 x 10^-5 = [Ag+]² x 0.142

[Ag+]² = 9.85 x 10^-5

[Ag+] = 0.0099 M

Since 2 moles of silver is got from each silver sulphate the solubility of silver sulpate in 0.142 M ammonium sulfate is 0.0099/2 = 0.00496 M

queen_honey_blossom answered 1 year ago

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