Part A Calcium chromate, CaCrO4, has a K sp value of 7.10×10 4 . What happens wh

Question

Part A

Calcium chromate, CaCrO4, has a Ksp value of 7.10×104 . What happens when calcium and chromate solutions are mixed to give 2.00×102M Ca2+ and 3.00×102M 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 1.00×102 M in the chloride ion, Cl? Ksp for lead(II) chloride is 1.17×105 .

Express your answer with the appropriate units.

Explanation / Answer

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

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