# Molar Solubility Questions

1. Aug 6, 2013

1. The problem statement, all variables and given/known data
I have worked these two problems to the best of my knowledge and I keep getting the wrong answer (I'm not sure what the right answer is). Here are the questions:

1. At a certain temperature the solubility of barium chromate (BaCrO4) is 1.9×10-5 mol/L. What is the Ksp of BaCrO4 at this temperature?
2. What is the molar solubility of silver phosphate (Ag3PO4) in water? Ksp = 1.8×10-18.

2. Relevant equations

The relevant equations are Ksp=([products]/[reactants])

3. The attempt at a solution
For question 1: BaCrO4⇔Ba2+ + CrO42-
Ksp=[Ba2+][CrO42-]/1 (since it is a solid)
Ksp=[x][x]=x2
Ksp=[1.9x10-5]2=3.61x10-10 = 3.6x10-10

For question 2: Ag3PO4⇔3Ag+ + PO43-
1.8x10-18 = ([Ag+]3[PO43-])/1
1.8x10-18 = [3x]3[x]
1.8x10-18 = [27x4] then divide both sides by 27
6.6666667x10-20 = x4 then do the fourth root on both sides
x = 1.6068568x10-5 = 1.6x10-5

I would very much appreciate any insight into this. Thank you!

2. Aug 6, 2013

### Saitama

Your approach looks fine to me unless you have done some calculation mistake.

3. Aug 6, 2013

It is an online homework, like a quiz, that we are allowed multiple attempts on. It doesn't give me the correct answer. I'll email the prof. Thanks!

4. Jan 4, 2017

### drusman313

where is this coming from??? [27x4]

5. Jan 5, 2017

### Staff: Mentor

Which part of $(3x)^3x = 27x^4$ is unclear to you? This is a pretty basic algebra.