Question #0a80d

• #46:

The first step to solving molar solubility problems is always to write down the balanced chemical equation of what's going on.

This would be:
`Ba_3(PO_4)_2 (s) rightleftharpoons 3Ba^(2+) (aq) + 2PO_4^(3-) (aq)`

Therefore the expression for `K_(sp)` would be:

`K_(sp) = [Ba^(2+)]^3[PO_4^(3-)]^2`

Setting up an ICE table would give you:

`K_(sp) = (3x)^3(2x)^2 = 108x^5`

Note: Watch your parentheses when simplifying these expressions

Now, the jump to take with all such problems is to realize that `x` is your molar solubility. In the ICE table, it represents how much of your solid dissolved, and therefore is the molar solubility.

You're given a molar solubility, so all you need to do is plug and chug:

`K_(sp) = 108(1.4 * 10^-8)^5 = 1.1 * 10^-38`

• #47:

Again, same process of setting up the expression:

`Ag_2CrO_4 (s) rightleftharpoons 2Ag^(+) (aq) + CrO_4^(2-) (aq)`

`K_(sp) = [Ag^(+)]^2[CrO_4^(2-)]`

By ICE Table:
`K_(sp) = (2x)^2(x) = 4x^3`

The `K_(sp)` value for this solid is `9.0 * 10^-12.` Hence:

`4x^3 = 9.0 * 10^-12`
`x = 1.3 * 10^-4 M`

• #48:

This problem is exactly the same process as the previous one, except with a different reaction:

`PbI_2 (s) rightleftharpoons 2I^(-) (aq) + Pb^(2+) (aq)`

`K_(sp) = [Pb^(2+)][I^(-)]^2`

By ICE Table:
`K_(sp) = (x)(2x)^2 = 4x^3`

The `K_(sp)` value for this solid is `7.1×10^(–9).` Hence:

`4x^3 = 7.1×10^(–9)`
`x = 0.0012 M`

To better understand what is exactly going on here, I'd recommend you watch these two videos.

Video 1 (Introduction to `K_(sp)`)
Video 2 (Examples)

Hope that helps :)