Question

When equal volumes of each solution are mixed, which combination will produce a precipitate of PbI2? (K sp of PbI2 = 8.5 × 10−9)

4.0 × 10−3 M Pb(NO3)2 and 2.0 × 10−3 M NaI

4.0 × 10−2 M Pb(NO3)2 and 8.0 × 10−5 M NaI

2.0 × 10−3 M Pb(NO3)2 and 4.0 × 10−2 M NaI

2.0 × 10−3 M Pb(NO3)2 and 2.0 × 10−4 M NaI

Answer 1

For first option:

We express the equilibrium of precipitation:

Pb⁺² + 2I^- <=> PbI₂

The ion product will be:

Ion product = [Pb⁺²][I^-]²

Concentration of [Pb⁺²] (assuming equal volumes):

C₁V₁ = C₂V₂

0.004M * x = C₂ * 2x

C₂ = 0.004/2 = 0.002 M

Concentration of I^-:

C₁V₁ = C₂V₂

0.002M * x = C₂ * 2x

C₂ = 0.002/2 = 0.001 M

Ion product = [0.002] * [0.001]² = 2 x 10^-9

As ion product is less than Ksp, no precipitation will occur.

Applying same procedure for all options, we get the following ion products:

Option 2: [0.02]*[0.00004]² = 3.2 x 10^-11

Option 3: [0.001]*[0.02]² = 4 x 10^-7

Option 4: [0.001]*[0.0001]² = 1 x 10^-11

As option 3 is the only option greater than the Ksp, in option 3 we'll have a precipitate.