Question

The Ksp of silver chromate is 1.20E-12. Calculate the


molar solubility (M) (Tol: ± 1E-007)
concentration of silver ion (M) (Tol: ± 1E-006)
concentration of chromate ion (M) (Tol: ± 1E-007)

Answer 1

@matt101

Answer 2

Molar solubility is how much of a substance is required to saturate a solution. Calculate this using the equation for Ksp:

\[K_{sp}=[Ag^+]^2[CrO_4^{2-}]\]
There is 1 mole of CrO4(2-) per mole of AgCrO4, so we will set its concentration to x (the molar solubility), which makes the concentration of Ag(+) equal to 2x (since there are 2 moles of Ag(+) per mole of AgCrO4). Now plug these values into the Ksp equation to solve for the molar solubility!

\[1.20 \times 10^{-12}=(2x)^2x\]\[x=6.7 \times 10^{-5}\]

So 6.7E-5 M is your molar solubility.

The concentration of your silver ion is going to be (2x)^2, and the concentration of your chromate ion is going to be x. You can do both of these calculations based on the Ksp equation above!

Answer 3

I'm getting molar solubility and chromate ion concentration right, but the concentration of the silver ion shows as incorrect @matt101

Answer 4

Sorry, just noticed my typo above. The concentration of silver ions is JUST 2x (no squaring). The squaring is what you do to the concentration as part of the Ksp equation.

So the concentration you should get is 13.4E-5 M = 1.3E-4 M!

Answer 5

Why is the value of it 2x, if its only one mol @matt101

Answer 6

x is your molar solubility. This means that it's your concentration of your WHOLE SOLID in solution after one mole of it has been introduced. However, this is not the same concentration as the final concentration of each ion in your solid.

Remember though that when an ionic solid dissolves, it dissociates - the ions it's made of split apart. Here we have Ag2CrO4. Every mole of Ag2CrO4 actually contains TWO MOLES of Ag(+) and one mole of CrO4(2-). So whatever the final concentration of Ag2CrO4 as a whole is (which is what the molar solubility is referring to), the concentration of Ag(+) is DOUBLE that, because each unit of Ag2CrO4 contains two of these ions. Similarly, since each unit of Ag2CrO4 contains only one CrO4(2-), the final concentration of CrO4(2-) in solution will be equal to the concentration of Ag2CrO4 as a whole - this is why it's concentration happens to be equal to the molar solubility.

Does that make more sense?