How do you know what is a "pure" solid or liquid?

[dedicate]

I was just doing problem 95 in EK chem, and I came across PbCl2. I didn't include Pb in the Ksp because I figured it was a "solid"... but I got it wrong.

What exactly is a pure solid or liquid? I don't fully understand the idea of [a pure solid's activity is 1] so you omit it. Obviously I know when given a rxn and it has (s) or (l) you omit it, but in this case, you just have a compound that you know dissolves to some extent.

If it was NaCl I am almost certain the Ksp is just Cl-, but how is Na+ a pure solid and Pb2+ is not? I don't know why of course lol, other than Na+ is always talked about.

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[dedicate]

Over 50 views and no reply! Someone must be able to shed some light on this!

 

[physics junkie]

The activity of a solid is 1 by definition. Ksp is called the solubility product because it is literally the product of the solubilities of the ions in moles per liter. The solubility product of a salt can therefore be calculated from its solubility, or vice versa.

I think you just need to memorize. EK general chem has a shorter list of the following in the first 3 chapters(I think):

Rules
Salts containing alkali metal cations or ammonium are soluble.
Salts containing nitrate, acetate chlorate or perchlorate anions are soluble.
Salts containing chloride, bromide or iodide are soluble. (Except for Cu+, Ag+, Hg2+2 and Pb+2)
Salts containing sulfate or sulfite are soluble. (Except Ca+2, Ba+2, Sr+2 and Pb+2)
Oxide and hydroxide salts are insoluble. (Except with group IA cations and ammonium.)
Sulfide and hydrogensulfide sals are insoluble. (Except with group IA cations and ammonium.)
Carbonate and chromate salts are insuluble. (Except with group IA and ammonium cations and ammonium.)

Edit: Actually I have no idea. To me this predicts that PbCl2 shouldn't be included.

 

[loveoforganic]

I'm a bit confused too, if the answer isn't Ksp = [Pb2+][Cl-]^2

 

[dedicate]

This was the question:

If the solubility of PbCl, is equal to x, which of the following expressions will be equal to the solubility product for PbCl2?

 

[sleepy425]

If it was NaCl I am almost certain the Ksp is just Cl-, but how is Na+ a pure solid and Pb2+ is not? I don't know why of course lol, other than Na+ is always talked about.

what???? no. for NaCl, the Ksp is [Na+][Cl-]. How do you get this? well, look at the solubility equilibrium:

NaCl (s) <---->Na+ (aq) + Cl- (aq)

The Ksp is the equilibrium constant for this equilibrium.

For PbCl2,

PbCl2 (s) <----> Pb2+ (aq) + 2Cl- (aq)

so the Ksp=[Pb2+][Cl-]^2

That's it, no exceptions, no complications, nothing fancy, a solid salt will dissociate in water into its corresponding ions. The ions will be in solution, the remaining solid will not be in solution.

 

[sleepy425]

The activity of a solid is 1 by definition. Ksp is called the solubility product because it is literally the product of the solubilities of the ions in moles per liter. The solubility product of a salt can therefore be calculated from its solubility, or vice versa.

I think you just need to memorize. EK general chem has a shorter list of the following in the first 3 chapters(I think):

Rules
Salts containing alkali metal cations or ammonium are soluble.
Salts containing nitrate, acetate chlorate or perchlorate anions are soluble.
Salts containing chloride, bromide or iodide are soluble. (Except for Cu+, Ag+, Hg2+2 and Pb+2)
Salts containing sulfate or sulfite are soluble. (Except Ca+2, Ba+2, Sr+2 and Pb+2)
Oxide and hydroxide salts are insoluble. (Except with group IA cations and ammonium.)
Sulfide and hydrogensulfide sals are insoluble. (Except with group IA cations and ammonium.)
Carbonate and chromate salts are insuluble. (Except with group IA and ammonium cations and ammonium.)

Edit: Actually I have no idea. To me this predicts that PbCl2 shouldn't be included.

So these are solubility rules. These are general solubility patterns used to tell if a salt is soluble in water. However, there is no such thing as something being completely insoluble in water. Every molecule has a certain degree of solubility in water, even if it is extremely small.

So, what is solubility? It is an equilibrium process between species in solution, and species not in solution (solids usually, although it could be a liquid or a gas).

Ok, so what does this equilibrium look like (in general):

Let's say we have a molecule X:

X (s) <----> X (aq)

The Ksp is just the equilibrium constant for this equilibrium, and since solids have activities of 1, the Ksp is just the concentration of X (aq). So that is the simplest case.

Most salts tend to ionize when they go into solution, which is why we commonly see a solid, which is one species, dissociate into two (or more) ions when they are in solution. However, in that case:

X (s) <----> Y (aq) + Z (aq) + W (aq) + ...

where Y, Z, and W are the different ions that X dissociates into. So our solubility equilibrium is the product of all of the aqueous species (and of course, raised to the power of their coefficient like usual).

So the only thing those solubility rules tell you is whether Ksp is high (high solubility) or low (low solubility).

 

[aevea]

I was just doing problem 95 in EK chem, and I came across PbCl2. I didn't include Pb in the Ksp because I figured it was a "solid"... but I got it wrong.

What exactly is a pure solid or liquid? I don't fully understand the idea of [a pure solid's activity is 1] so you omit it. Obviously I know when given a rxn and it has (s) or (l) you omit it, but in this case, you just have a compound that you know dissolves to some extent.

If it was NaCl I am almost certain the Ksp is just Cl-, but how is Na+ a pure solid and Pb2+ is not? I don't know why of course lol, other than Na+ is always talked about.

In PbCl₂, the Pb isn't Pb⁰ but Pb²⁺. So, after solvation, some (not much) of the Pb²⁺ is in solution, just like the Cl^-, and therefore should be included in the K_sp calculation. Undissociated PbCl₂ on the other hand will be a pure solid precipitate on the bottom of the vessel and is therefore ignored in the K_sp.

In NaCl it's the same thing, but the math is easier: K_sp=[Na⁺][Cl^-]. So the solubility is equal to the amount of Na⁺ or Cl^- in solution and the K_sp is equal to the square of the solubility.

This was the question:

If the solubility of PbCl2, is equal to x, which of the following expressions will be equal to the solubility product for PbCl2?

So, here the math is trickier. Start out with the K_sp=[Pb²⁺][Cl^-]². The square on the chloride ion is from the coefficient in the balanced equation. Now, if the solubility of PbCl₂ is x that means that for every x moles of PbCl₂ that dissolves you'll get x moles of Pb²⁺ and 2x moles of Cl^- in solution. So, the K_sp=[x][2x]²=x(4x²)=4x³. It seems like your counting the doubling of chloride ion twice, but that's how you do K_sp's.