# Solubility - solubility product

1. May 5, 2012

### carle

Hello! Two questions that I would like to get some help with. I'm translating these so I hope I get it right.

1. The solubility of cobalt(II)hydroxide is 5.4*10^-6 moles/dm^3 in water.
i) What is the solubility product?
ii) What is the solubility of cobalt(II)hydroxide if pH is buffered to 10.43?

In i) I've calculated Ks to 6.3*10^-16 which is correct. The answer in ii) is 8.6*10^-9.

2. pH in a saturated magnesium hydroxide solution is 10.52. Calculate the solubility product for magnesium hydroxide.

The answer is 1.8*10^-11.

I've tried to solve them for a long time but I can't get it right. I need some help on how to approach problems like these.

2. May 5, 2012

### Staff: Mentor

Show how you tried to solve, we will start from there.

3. May 5, 2012

### carle

1
i) The reaction will be Co(OH)2 ⇔ Co2+ + 2OH- so Ks = [Co2+][OH-]2 which gives (5.4*10-6)*(2*5,4*10-6)2 = 6,3*10-16.

ii) pH is 10.43 so pOH = 3.57 and [OH-] = 10-3.57. I think that I need to do some kind of schedule between the mole ratios, because I can see that the concentration ratio between Co2+ + 2OH- is 1:2. But I don't know how to proceed.

2

The reaction will be Mg(OH)2 ⇔ Mg2+ +2OH-. We know that pH = 10.52 so [OH-] = 10-3.48. Ks = [Mg2+][OH-]2, but once again I don't know how to proceed with the concentration ratios.

4. May 5, 2012

### Staff: Mentor

Concentration ratios are important only when the salt itself is the only source of all ions. In both questions here it is enough to solve Ksp for the unknown and to plug known concentration in.

5. May 5, 2012

### carle

Hm.. but in 2) [Mg2+ is unknown, how do I calculate that?

6. May 5, 2012

### Staff: Mentor

Exactly as I told you. [Mg2+] is the only unknown. You know [OH-], you take Ksp from tables, plug and chug.

I feel like you are missing what the Ksp is and how it works. If the solution is saturated product of concentrations (taken to correct powers) equals Ksp - so if you know Ksp and you are given concentrations of all ions but one, you can calculate concentration of this one ion - always. When you deal with solubility you are not given concentrations, but you know from the dissolution stoichiometry how they related to each other, so you can calculate them all. But these are two different cases.