Question

Part A

Calcium chromate, CaCrO4, has a Ksp value of 7.10×10⁻⁴ . What happens when calcium and chromate solutions are mixed to give 2.00×10⁻²M Ca2+ and 3.00×10⁻²M CrO42−?

A) A precipitate forms because Q>Ksp.

B) A precipitate forms because Q<Ksp.

C) No precipitate forms because Q>Ksp.

D) No precipitate forms because Q<Ksp.

Part B

What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbCl2 from a solution that is 1.00×10−2 M in the chloride ion, Cl−? Ksp for lead(II) chloride is 1.17×10⁻⁵ .

Express your answer with the appropriate units.

Answer 1

Q A Solution :

Given data

Ksp = 7.10*10^-4

[Ca^2+] = 2.00*10^-2 M

And [CrO4^2-] =3.00*10^-2 M

Lets calculate the Qsp using the given concentrations

Qsp = [Ca^2+][CrO4^2-]

Lets put the values in the formula

Qsp = [2.00*10^-2][ 3.00*10^-2]

Qsp = 6.0*10^-6

The calculated value of the Qsp is smaller than the value of the ksp

Therefore correct statement is option D

That is

No precipitate forms because Q<Ksp

Q B Solution

Given data

PbCl2 ksp = 1.17*10^-5

[Cl^-] = 1.00*10^-2 M

[Pb^2+] = ?

Dissociation equation for the PbCl2 is as follows

PbCl2 ----- > Pb^2+ + 2Cl^-

                          X           2x

Ksp equation is as follows

Ksp = [Pb^2+ ] [Cl^-]²

Lets put the values in the formula

1.17*10^-5 = [x][ 1.00*10^-2]²

x=1.17*10^-5 /[ 1.00*10^-2]²

x= 0.117 M

therefore the concentration of the Pb^2+ needed to start precipitation = 0.117 M