Calculate the molar solubility of Ag2SO4 (Ksp = 1.5 x 10 ^-5) a) in pure water b

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Explanation / Answer

Ag2SO4 ---> 2Ag+ + SO4-2 one mole gives two moles of Ag+ and one mole of SO4-2 solubility product Ksp = [Ag+]^2 * [SO4-2] => Ksp = (2S)^2 * S = 4 *S^3 => 1.5 * 10^-5 = 4 * S^-3 solving u get S = 0.0155 hence molar solubility of Ag2SO4 in water is 0.0155 now if we have 0.22M Na2SO4 Ksp = [Ag+]^2 * [SO4-2] since Na2SO4 dissociates completely [SO4-2] = 0.22 M hence 1.5 * 10^-5 = 4 * S^2 * 0.22 on solving u get S = 0.0026 Hence solubility of Ag2SO4 in 0.22M Na2SO4 is 0.0026

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