chem questions?

Question

1. Determine the molar solubility of PbI2. The Ksp values for PbI2 is 8.7 x 10^-9.

2. Determine the molar solubility of 100 mL of a solution of PbI2 to which 0.01 mole of lead nitrate, Pb(NO3)2 has been added. Assume the total volume remains at 100 mL.

3. The Ksp valu for calcium carbonate is 4.5 x 10^-9. Calculate the solubility of calcium carbonate in g/L.

If you could explain the steps I would really appreciate it so I can figure out the others on my own.

Answer 1

The equilibrium is :
PbI2 <----> Pb2+ + 2I-
let x = mole/L of PbI2 that dissolve. This gives us x mol/L Pb2+ and 2x mol/L I-
8.7 x 10^-9 = ( x) (2x)^2 = 4x^3
x = molar solubility = 0.0013 M

[Pb2+] = x + 0.01
[I-] = 2x
8.7 x 10^-9 = (x+ 0.01) (2x)^2
x = 0.00047 M ( moles that dissolve in 1 L). To get moles in 100 mL =>0.000047 moles in 100 mL

CaCO3 <----> Ca2+ + CO32-
Ksp = 4.5 x 10^-9 = x^2
x = molar solubility = square root 4.5 x 10^-9 = 6.7 x 10^-5 M
Grams dissolved in 1 L = 6.7 x 10^-5 mol/L x 100 g/mol =
6.7 x 10^-3 g

Answer 2

PbI2 <===> Pb++ + 2I-

1. Ksp = [Pb++][I-]^2 = 8.7 x 10^-9

[Pb++] = The molar solubility of PbI2. Let [Pb++] be called x.

[I-] = 2[Pb++]

Ksp = (x)(2x)^2 = 4x^3 = 8.7 x 10^-9

x^3 = 2.2 x 10^-9

x = 1.3 x 10^-3 moles/L (Answer to 1.)

2. 0.01molPb(NO3)2/100mL = 0.1molPb(NO3)2/L (molarity)

Because PbI2 <===> Pb++ + 2I-, [I-] = 0.5 the molar solubility. Let [I-] be called x.

Ksp= [Pb++][I-]^2 = 8.7 x 10^-9

(0.1)x^2 = 8.7 x 10^-9

x^2 = 8.7 x 10^-10

x = 2.94 x 10^-5

But that's only 1/2 the molar solubility, which is 1.5 x 10^-5moles/L (The answer to 2.)

We used 0.1M for [Pb++], because the added Pb(NO3)2 is so much greater than [Pb++] from molar solubility that the latter doesn't count.

3. Use the same method as in 1.

Answer 3

Solubility Of Pbi2

Answer 4

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