Question

Select the false statement below regarding PbI2 (Ksp = 7.9 x 10−9) and PbCO3 (Ksp = 7.4 x 10−14).

  

A. PbCO3 is more soluble in 1.0 M  Pb(NO3)2 than in pure water

B. PbCO3 is more soluble in 1.0 M HCl than in pure water

C. the molar solubility of PbI2 can be calculated by taking the cube root of (Ksp/4)

D. PbI2 is more soluble in 1.0 M NaOH than in pure water (Kf of Pb(OH)3− = 8 x 1013)

E. PbI2 is less soluble in 1.0 M NaI than in pure water

A and D are wrong

Answer 1

A. PbCO3 is more soluble in 1.0 M  Pb(NO3)2 than in pure water

Note that, by definition, Nitrate ions, NO3- are soluble, thereofre Pb(NO3)2 is much more soluble than PbCO3, this must be false statement

B. PbCO3 is more soluble in 1.0 M HCl than in pure water

PbCO3(s) -> PB+2 + CO3-2(aq)

if we add H+, then

H+ + CO3-2(aq) --> HCO3-(aq)

meaning that the shift of solid PbCO3 must go from left to right, in order to counterbalance the shift

so this must increase PbCO3 in solution, therefore, TRUE

C. the molar solubility of PbI2 can be calculated by taking the cube root of (Ksp/4)

True,

math proof:

PbI2 --> Pb+2 + 2I- therefore

Ksp = [PB+2][I-]^2

Ksp = (S)(2S)^2

Ksp = 4*S^3

S = (Ksp/4)^(1/3)

D. PbI2 is more soluble in 1.0 M NaOH than in pure water (Kf of Pb(OH)3− = 8 x 1013)

False, 1M is not enough to form Pb(OH)3- complex, it will form Pb(OH)2(s) as well

E. PbI2 is less soluble in 1.0 M NaI than in pure water

True, since

PbI2 --> Pb+2 + 2I-

if we add NaI --> Na+ + I- , thereofre, I- increases

the shift goes from aqueous Pb+2 and I- to PbI2(s) solid, decreasing solubility