Question

When 22.0 mL of a 2.08×10^-4M ammonium iodide solution is combined with 18.0 mL of a 5.92×10^-4M lead acetate solution does a precipitate form? _____ (yes or no)

For these conditions the Reaction Quotient, Q, is equal to _____

Answer 1

Number of moles of NH4I , n = Molarity * volume in L

= 2.08*10-4 M * 22.0mL*10-3L/mL

= 4.576*10-6 moles

Number of moles of lead acetate ,n' = Molarity * volume in L

= 5.92*10-4 * 18.0mL*10-3L/mL

= 1.0656*10-5 moles

The balanced reaction is : 2NH4I (aq) + Pb(CH3COO)2 (aq) ----> PbI2(s) + 2CH3COONH4(aq)

After combining the total volume of the solution is = 22.0mL+18.0mL = 40.0 mL = 0.040 L

[Pb(CH3COO)2 ] = number of moles of Pb(CH3COO)2 / total volume

= 1.0656*10-5 moles / 0.040 L

= 2.664*10-4 M

So [Pb2+] = [Pb(CH3COO)2 ] = 2.664*10-4 M

[NH4I] = number of moles of NH4I / total volume

= 4.576*10-6 moles / 0.040 L

= 1.144*10-4 M

[I- ] = NH4I] = 1.144*10-4 M

1 mole of PbI2 contains 1 mole of Pb2+ & 2 moles of I-

So total concentration of [I- ] = 2 * 1.144*10-4 M = 2.288*10-4 M

Reaction Quotient , Q = [Pb2+][I- ]2

= (2.664*10-4 M) (2.288*10-4 M )2

= 1.39*10-11

The solubility product constant of PbI2 , Ksp = 8.7*10-9

Since Q < Ksp so precipitate will not be formed, dissolution takes place.