0:09

Now very commonly what students will try to do when they come across a problem like

Â this is to react these two as reactants and try to come up with a product, but

Â that is not the way you work these problems.

Â What you will want to do is write the reaction as you did in part A.

Â The silver chromate dissolving in solution

Â to give you two silver ions plus chromate.

Â 0:37

And the change or the difference in part B versus part A comes in this I line.

Â You will put some of the solid into the beaker where there's already

Â some solution present of sodium, or some ions present of sodium and chromate.

Â So there's not any silver ions present but

Â there is 2.6 times 10 to the minus 3 molar of the chromate present.

Â 1:02

And so I'll put that number here saying it's this number.

Â Now word of caution.

Â The reason it's the same number is because there is

Â one of the chromate in sodium chromate.

Â So the concentration of chromate is the same as this.

Â But if there were two chromates in the formula, which there isn't, but

Â if there were, then I would need to double that value that they're giving me here

Â before I put it into the table,

Â because that would be the concentration of the ions in solution.

Â 1:32

So we will use minuses here because that's how much

Â of the sodium chromate will dissolve.

Â That's the molar solubility.

Â I will add 2s here.

Â I will add s over here.

Â That will still give me some solid in solution, I mean in the container,

Â I have 2s for that, and I have 2.6 x 10 to the minus 3 plus s for this.

Â Now, I will always want to write the Ksp expression it's

Â the silver ion concentration squared.

Â I'm looking at the reaction, I'm doing the powers a lot of times students try to skip

Â this step to save time, but

Â then they invariably forget to work with those coefficients.

Â Then I will look at the silver concentration, and that was 2s, and

Â it's squared, and then I will look at the chromate concentration, which I see here,

Â 2.6 times 10 to the minus 3 plus s, so all of this is multiplied.

Â Now we're going to assume that S is very small compared to the other values,

Â and ignore this term for simplicity sake.

Â We'll make sure we go back and check that it was a valid assumption.

Â So all of this is going to be equal to the number

Â of ksp which is given in the problem.

Â 1.2 times 10 to the minus 12.

Â And so I will multiply the 2 times 2 that will be 2 squared and

Â then I'm going to multiply that to the 2.6 x 10 to the -3.

Â And that is going to give me 1.04 times 10 to the minus 2 times s squared.

Â So then I will divide both sides by the 1.04 times 10 to the minus 2,

Â 3:25

So then I can take the square root of both sides, and that will give me s, and

Â the value for s is 1.0 times 10 to the minus 5.

Â So then we check our assumption, and if we take this value and

Â subtract it from the 2.6 times 10 to the minus 3,

Â we will still get 2.6 times 10 to the minus 3 or add it to it, add it,

Â it won't change the values, so that's a good assumption for us.

Â And so, this is the molar solubility of silver chromate

Â In a solution that contains sodium chromate.

Â Now let's see if it makes sense.

Â The solubility of the silver chromate in pure water,

Â that was determined in part A,

Â was 6.54 times 10 to the minus 5.

Â 4:19

Okay, and this is 1 times 10 to the minus 5, so the molar solubility

Â has decreased by the addition of this small amount of the sodium chromate,

Â so it is supporting what we know about a common ion situation.

Â Common ions always decrease the solubility of an insoluble salt.

Â