# Solubility calculations

Hello,

I am a bit confused re. solubility calculations.

Calculate the solubility of Pb(OH)2 at pH 10. Setting up the expression for Ksp:

Ksp = [Pb2+][OH-]2 = 8* 10^-17 (Ksp value from SI chemical data)

pOH = 14 - 10 = 4, i.e. [OH-] = 10-4.

8*10-17 = [Pb2+][OH-]2. Solving for [Pb2+] we get that [Pb2+] = 8*10-9.

My question is: why couldn't we use the relationship between the concentrations of Pb2+ and OH- like we do with ICE tables, to get the result? I mean, couldn't we just divide the concentration of OH- by 2 and get the concentration of Pb2+..?

BvU
Homework Helper
Doesn't the bulk of the [OH] come from water ?

Borek
Mentor
Compare concentration of OH- already present pH 10 with the concentration of OH- that would be produced just by Pb(OH)2 dissolution. Does the addition of OH- from the dissolution matter?

So, at pH 10 the [OH-] concentration will be 10-4. Whereas the concentration from *just* the Pb(OH)2 dissolution would be 5.42*10-6, right? (At neutral pH) But due the significantly higher concentration of [OH-] from the pH 10, we have to include that in the calculations?

James Pelezo
Gold Member
You can use the ICE table to show the process. It may help to better understand this problem by considering the difference between solubility of the Pb(OH)2 in, say DI water at pH = 7 and OH- treated water at pH = 10. Set up the ICE table and in the 'Initial' row enter '0' Molar for the Pb+2 concentration and 10-4 M for the OH- concentration as the common ion in the solution. (pOH = 14 - pH => pOH = 14 - 10 = 4 => [OH-] = 10-4) Fill in the rest of the table with variables per standard practice. Since the OH- is already present at 10-4 & additional OH- from further ionization of Pb(OH)2 is negligible, equilibrium row of the ICE table would be 'x' for [Pb+2 and 10-4 for [OH-. Apply to the Ksp expression for Pb(OH)2 and solve for x = [Pb+2] at pH = 10. Compare this with solubility of Pb(OH)2 in DI water ( for 1:2 ionization in DI water S = (Ksp/4)1/3. You should see a decrease in solubility on the order of 102.

Last edited:
James Pelezo
Gold Member
Hello,

I am a bit confused re. solubility calculations.

Calculate the solubility of Pb(OH)2 at pH 10. Setting up the expression for Ksp:

Ksp = [Pb2+][OH-]2 = 8* 10^-17 (Ksp value from SI chemical data)

pOH = 14 - 10 = 4, i.e. [OH-] = 10-4.

8*10-17 = [Pb2+][OH-]2. Solving for [Pb2+] we get that [Pb2+] = 8*10-9.

My question is: why couldn't we use the relationship between the concentrations of Pb2+ and OH- like we do with ICE tables, to get the result? I mean, couldn't we just divide the concentration of OH- by 2 and get the concentration of Pb2+..?

Check your Ksp value... I found Ksp to be different at => https://users.stlcc.edu/gkrishnan/ksptable.html

Borek
Mentor
So, at pH 10 the [OH-] concentration will be 10-4. Whereas the concentration from *just* the Pb(OH)2 dissolution would be 5.42*10-6, right? (At neutral pH) But due the significantly higher concentration of [OH-] from the pH 10, we have to include that in the calculations?

It is not like we to include it, we actually have to base calculations on it. Concentration of OH- already present is two orders of magnitude higher than what would be added from the Pb(OH)2 dissolution, so the effect of the dissolution can be rather safely ignored, while the effect of the ions present can not.

James Pelezo
Gold Member
You are exactly right ... I am just offering a 'perspective' approach to solving/visualizing the problem by 'assuming' the OH- ions are already present as a common ion and Pb(OH)2 is added into that solution. The high OH- concentration would of course inhibit ionization of the Pb(OH)2 as opposed to ionization in pH = 7 water. Calculation shows ~100x drop in solubility. at [OH-] ~ 10-4M levels. I often ask students to consider 'How much of the compound/salt would deliver ions into solution with vs without the common ion'. Of course, your right in that the ions are of primary consideration and the lower the concentration of ions delivered into solution the lower the solubility; i.e., as you say 'safely ignored'. Have a great day! :-)

Thanks, both of you!

I have another question re. solubility:

You have a solution of 0.1M AgNO3 and 0.09M KI, and AgI is precipitated. What is [Ag+] and [I-] in the solution after precipitation?

Given the higher concentration of AgNO3 I understand there must be 0.1 - 0.09 = 0.01 M [Ag+] left in the solution. But what about the [I-? Isn't all this consumed in the precipitation of AgI?

Borek
Mentor

Isn't all this consumed in the precipitation of AgI?

Imagine putting some solid AgI in the 0.01 M solution of Ag+. What would happen?

Some [I-] would react and precipitate AgI whereas the rest would just stay as Ag+ and I- ions..?

Borek
Mentor
There were no I- ions in the case I have suggested - please reread what I wrote.

Ok, but won't some of the added AgI dissolve in your case? That was what I meant by I-.

Borek
Mentor
Ok, but won't some of the added AgI dissolve in your case? That was what I meant by I-.

Yes, some would dissolve, but it would not precipitate any additional Ag+ - quite the opposite, when AgI dissolves concentration of Ag+ will (very slightly) increase. But you are right that some I- will show in the solution. Can you calculate (using Ksp) how much?

Yes, some would dissolve, but it would not precipitate any additional Ag+ - quite the opposite, when AgI dissolves concentration of Ag+ will (very slightly) increase. But you are right that some I- will show in the solution. Can you calculate (using Ksp) how much?
Isn't (bold part) a contradiction? I know how I can calculate the amount of I-, knowing the Ksp of AgI and the concentration of [Ag+] = 0.01M, and the result is a veery small amount of [I-]. What I don't understand is where this [I-] comes from, because I thought all of it was consumed in the precipitation of AgI.

BvU