Lead(II) chromate has a K sp of 2.0 x 10 -16 . Exactly 4.0 mL of 0.0040 M lead(I

ID: 505138 • Letter: L

Question

Lead(II) chromate has a K sp of 2.0 x 10 -16 . Exactly 4.0 mL of 0.0040 M lead(II) nitrate is mixed with 2.0

mL of 0.00020 M sodium chromate.

a. Write the precipitation reaction (net ionic) and the equation for the solid precipitate

dissociating.

b. Show the K sp expression for this solid precipitate dissociating.

c. Will a precipitate form? Show calculations to support your answer.

d. What would be the effect on the solubility equilibrium system if concentrated potassium

chromate solution is added?

a) Pb2+(aq) + CrO42-(aq) -------> PbCrO4(s)

PbCrO4(s) <----------> Pb2+(aq) + CrO42-(aq)

b) Ksp = [Pb2+][CrO42-]

Ksp = 2.0× 10^-16

c) Initial [Pb(NO3)2] =0.0040M

Final volume = 6ml

Dilution = 6/4= 1.5time

Final [ Pb(NO3)2] = 0.0040/1.5 = 0.00266M

Final [Pb2+] = 0.00266M

Initial [Na2CrO4] = 0.0002M

Dilution = 6/2= 3

Final [ Na2CrO4]= 0.0002/3= 6.66×10^-5

Final [CrO42-] = 6.66×10^-5M

Qsp of PbCrO4 = [Pb2+] [CrO42-]

=0.00266 × 6.66×10^-5

= 1.8 × 10^-7

Qsp > Ksp

So , precipitation of PbCrO4 will form

d)When concentrated K2CrO4 added, the solubility equilibrium will shift to left side and precipitate of PbCrO4 wil form

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