[Coruscantian No.66]

5. What is the molar solubility of PbI2 in a 0.10 M NaI solution? Ksp for PbI2 is 7.9*10^-9

i know how to do this question if the solvent is Water instead of NaI, does the nature of the solvent have any effect on the solubility equilibrium?  

4. The value of Ksp for AgCl is 1.8*10^-10, what is the molar solubility of AgCl in pure water?

My answer is 1.34*10^-5 mol/L, but what is molar solubility? is it the same as solubility?

thanks in advance

[kkrizka]

5. Water is still the solvent. Basically what solution means usually is that something (NaI in this case) is dissolved in water.

[Donaldson Tan]

5. What is the molar solubility of PbI2 in a 0.10 M NaI solution? Ksp for PbI2 is 7.9*10^-9

i know how to do this question if the solvent is Water instead of NaI, does the nature of the solvent have any effect on the solubility equilibrium?  

Ksp = [Pb2+][I-]²
let amount of PbI₂ dissolve per unit volume of solution be x
=> [Pb2+] = x
=>  [I-] = 0.1 + 2x
=> Ksp = x(0.1 + 2x)
Solve for x

4. The value of Ksp for AgCl is 1.8*10^-10, what is the molar solubility of AgCl in pure water?

My answer is 1.34*10^-5 mol/L, but what is molar solubility? is it the same as solubility?

Molar solubility is solubility in units of mole per unit volume.

[Borek]

[I-] = 0.1 + x

1. For so low Ksp you may safely assume that concentration of iodide doesn't change.

2. You are wrong. Think about the stoichiometry.

[Donaldson Tan]

2. You are wrong. Think about the stoichiometry.

Ksp = [Pb2+][I-]²
consider 1dm3 0.1M NaI solution
PbI₂ <-> Pb²⁺ + 2I^-
let amount of PbI₂ dissolve be x
=> [Pb2+] = x/V = x (since V = 1dm3)
every mole of PbI2 dissolves to produce 2 moles of iodide in solution.
=>  [I-] = 0.1 + 2x
Ksp = x(0.1 + 2x)

1. For so low Ksp you may safely assume that concentration of iodide doesn't change.

for low Ksp, x is usually small such that 0.1 >> x, so 0.1 + 2x is approximately 0.1. This means the K_sp equation reduces to K_sp = 0.1x

[markajane03]

Don't you have to square the quantity (0.1 + 2x) since there are 2 I- atoms for ever Pb2+ atom?  I get (x)(0.10 + 2x)^2 = 7.9 x 10^-9.  Can't you make the assumption that 0.10 + 2x = 7.9 x 10 ^-9?

[AWK]

2. You are wrong. Think about the stoichiometry.

Ksp = [Pb2+][I-]²
consider 1dm3 0.1M NaI solution
PbI₂ <-> Pb²⁺ + 2I^-
let amount of PbI₂ dissolve be x
=> [Pb2+] = x/V = x (since V = 1dm3)
every mole of PbI2 dissolves to produce 2 moles of iodide in solution.
=>  [I-] = 0.1 + 2x
Ksp = x(0.1 + 2x)

1. For so low Ksp you may safely assume that concentration of iodide doesn't change.

for low Ksp, x is usually small such that 0.1 >> x, so 0.1 + 2x is approximately 0.1. This means the K_sp equation reduces to K_sp = 0.1x

Shoud be
Ksp = x(0.1 + 2x)²
As Borek pointed out at the first approximation we can assume  0.1 + x = 0.1
Note, the real solubility will be higher by 20-30 % because of ionic strength effect