Question

determine the maximum amount (grams) of silver sulfate( Ag2SO4) that can be dissolved into 1.00 L of a 0.22 M AgNO3 solution given the Ksp of silver sulfate is 1.5x10^-5

Answer 1

AgNO3 here is Strong electrolyte

It will dissociate completely to give [Ag+] = 0.22 M

At equilibrium:

Ag2SO4 <----> 2 Ag+ + SO42-

   0.22 +2s s

Ksp = [Ag+]^2[SO42-]

1.5*10^-5=(0.22 + 2 s)^2*(s)

Since Ksp is small, s can be ignored as compared to 0.22

Above expression thus becomes:

1.5*10^-5=(0.22)^2*(s)

1.5*10^-5= 4.84*10^-2 * 1(s)^1

s = 3.099*10^-4 M

Molar mass of Ag2SO4 = 2*MM(Ag) + 1*MM(S) + 4*MM(O)

= 2*107.9 + 1*32.07 + 4*16.0

= 311.87 g/mol

Molar mass of Ag2SO4= 311.87 g/mol

s = 3.099*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 3.099*10^-4 mol/L * 311.87 g/mol

s = 9.7*10^-2 g/L

since volume is 1 L, mass dissolved = 9.7*10^-2 g

Answer: 9.7*10^-2 g