Question

 The molar solubility of lead phosphate in a 0.180 M potassium phosphate solution is _______  M.

 The maximum amount of lead sulfate that will dissolve in a 0.299 M ammonium sulfate solution is _______M.

 The molar solubility of lead phosphate in a 0.259 M lead nitrate solution is _______  M.

Answer 1

A.

K3PO4(aq) ---------> 3K^+(aq) + PO4^3- (aq)

0.18M -----------------------------------0.18M

Pb3(PO4)2(s) ---------------> 3Pb^2+(aq) + 2PO4^3-(aq)

--------------------------------------- 3x--------------- 2x+0.18

Ksp   = [Pb^2+]^3[PO4^3-]^2

1*10^-54 = (3x)^3(2x+0.18)^2

1*10^-54 = (3x)^3(0.18)^2       [ 2x+0.18 = 0.18 , 2x<<<<<0.18]

x   = 1.04*10^-18 M

molar solubility of lead phospahte is 1.04*10^-18M

B.

(NH4)2SO4(aq) -----------------> 2NH4^+(aq) + SO4^2- (aq)

0.299M ------------------------------- 2*0.299M----------- 0.299M

PbSO4(s) -------------> Pb^2+ (aq) + SO4^2- (aq)

----------------------------- x ------------------x+0.299

Ksp   = [Pb^2+][SO4^2-]

1.3*10^-8 = x*(x+0.299)

1.3*10^-8 = x*(0.299)          [ x + 0.299 = 0.299 , x <<<<0.299]

x = 4.35*10^-8

The maxium amount 4.35*10^-8 M

c.

Pb(NO3)2(aq) ------------> Pb^2+(aq) + 2NO3^- (aq)

0.259M --------------------------0.259M

Pb3(PO4)2(s) ---------------> 3Pb^2+(aq) + 2PO4^3-(aq)

--------------------------------------- 3x+0.259------- 2x

Ksp   = [Pb^2+]^3[PO4^3-]^2

1*10^-54 = (3x+0.259)^3(2x)^2

1*10^-54 = (0.259)^3(2x)^2      [ 3x+0.259   = 0.259 , 3x<<<<<0.259]

x   = 3.8*10^-27

The molar solubility = 3.8*10^-27 M