Question

To 100 ml of 0.1 M AgNO3 solution solid K2SO4 s added. The concentration of K2SO4 that shows precipitation is? (Ksp for Ag2SO4 = 6.4 * 10^-5 M)
(A) 0.1 M
(B) 6.4 * 10^-3 M
(C) 5.4 * 10^-7 M
(D) 6.4 * 10^-5 M

Answer 1

2 Ag NO3 (aq) + K2 SO4 (aq) ==> Ag2 SO4 (s) + 2 K NO3 (aq)

Silver Sulfate is a non soluble precipitate.  Ag NO3 is completely soluble.

Number of moles of AgNO3 in 0.1L of 0.1M solution: 0.1*0.1 = 0.01 moles.
Number of moles of K2SO4 reacting with 0.01 moles of AgNO3: 0.005 moles.

Ag2SO4  <==> 2 Ag⁺       +     SO4⁻²
insoluble           soluble
 1 - x                    2 x moles      x moles

Ksp = Solubility product = [ Ag⁺ ]² [ SO4⁻² ] = 6.4 * 10⁻⁵ 

Precipitate forms if Ionic (concentration) product IP is more than Solubility product Ksp.

We have [Ag⁺] = 0.1 M.     [ SO4⁻ ] = to find
     IP = [Ag⁺]² [ SO₄ ⁻] = 0.1²  [SO4⁻ ] = 0.01 [ SO4 ⁻ ]
        IP > Ksp

So   0.01  [ SO4⁻ ]  > 6.4 * 10⁻⁵ 
       [ SO4 ⁻ ]  > 6.4 * 10⁻³ M

In K2 SO4 there is only one Sulphate ion per molecule.  So if concentration of Potassium Sulfate is more than this , then precipitate will form.