Question

When silver sulfate (Ag2SO4) is dissolved in water at 25 °C, until saturation is reached, the Ag+ cation equilibrium concentration is 3.0 x 10 ^-5 M.
a) Determine the value of the solubility constant for Ag2SO4
(b) Determine the maximum amount (in grams) of AgNO3, a completely water-soluble salt, which can be dissolved in 3.2 L of a H2SO4 solution without precipitating Ag2SO4

Note: Actually the molarity of H2SO4 is not given in the exercise. Is it really necessary for the exercise resolution? If it is needed can you please pick a random molarity that you think fits the exercise. At least that way I can learn how to solve this kind of exercises. Thanks in advance!

Answer 1

Part (a).

Ag₂SO₄ <----> 2Ag⁺(aq) + SO₄^2-(aq)

Here, [Ag⁺] = 3*10^-5 M

Now, the solubility product constant (Ksp) of Ag₂SO₄ = [Ag⁺]²[SO₄^2-] = (2s)²*(s)

Where, s = molar solubility of Ag₂SO₄ = 3*10^-5 M/2 = 1.5*10^-5 M

Hence, Ksp = (3*10^-5)²*(1.5*10^-5) = 1.35*10^-14

Part (b).

Up to the value of [Ag⁺] = 3*10^-5 M, the precipitation Ag₂SO₄ will not take place.

i.e. The moles of Ag⁺ = 3*10^-5 mol/L * 3.2 L = 9.6*10^-5 mol

Therefore, the required mass of AgNO₃ = 9.6*10^-5 mol * 169.87 g/mol = 0.0163 g