Part A: A solution contains 7.68×10 -3 M iron(II) acetate and 5.92×10 -3 M zinc

Question

Part A: A solution contains 7.68×10-3 M iron(II) acetate and 5.92×10-3 M zinc nitrate.
Solid potassium hydroxide is added slowly to this mixture.

A. What is the formula of the substance that precipitates first?

B. What is the concentration of hydroxide ion when this precipitation first begins? [OH-]=____ M ?

Part B: A solution contains 1.41×10-2 M ammonium sulfide and 6.35×10-3 M sodium hydroxide.
Solid iron(III) nitrate is added slowly to this mixture.

A. What is the formula of the substance that precipitates first?

B. What is the concentration of iron(III) ion when this precipitation first begins?

[Fe3+]=______M?

Explanation / Answer

I will tell you how it's done part A) so you can have a guidance of how this kind of problem is done, and then, try to solve Part B by yourself. If you still have doubts, you can post that question again and I'll gladly answer it for you.

First, the solution contains the following compounds:
Fe(CH3COO)2 and Zn(NO3)2

When the KOH is added, it may form the following hydroxides:
Fe(OH)2 and Zn(OH)2

How do we know which will precipitate first? we need the values of the Ksp. The one with the lower (or smaller) value will be the one to precipitate first.

Ksp Fe(OH)2 = 2x10-15
Ksp Zn(OH)2 = 5x10-17

Therefore, the Zn(OH)2 will precipitate first. and the concentration of the OH- will be:
Zn(OH)2 --------> Zn2+ + 2OH-
s 2s

Ksp = [Zn][OH]2
5x10-17 = s * (2s)2
5x10-17 = 4s3
s = (5x10-17 / 4)1/3
s = 2.32x10-6 M
[OH-] = 2*2.32x10-6 = 4.64x10-6 M

Part B, you may solve it following the same procedure of here. Find the Ksp of the possible products, the lower one will be precipitating first, and then, calculate the concentration.

Hope this guidance helps.

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