Question

I did an experiment where I had 5 ml of MgCl2, then 5 ml NH4 solution (NH4OH (aq)) was added and it was observed that a precipitate formed; cloudy solution. After I added 1g of NH4Cl(s), the solution clears up (transparent).

2nd experiment:
There was 5ml Lead (II) Nitrate in a test tube, 5ml sodium oxalate was added to it, a precipitation forms; cloudy solution. The precipitate settled and the liquid above it was decanted. After, dilute nitrite acid was added some precipitate dissolved (still a bit cloudy).

Can someone explain which of the species are reacting to form/dissolve the precipitate and how it dissolves after in both experiments? Providing the equilibrium equations would help with the explanation. Thanks

Answer 1

I don't know what nitrite acid is so I can't help you with the second part. The first part is this.

Mg^2+ + 2OH^- ==> Mg(OH)2.
1) Ksp = (Mg^2+)(OH^-)

NH3 + H2O ==> NH4^+ + OH^-
2) Kb = (NH4^+)(OH^-)/(NH3)
[note: my NH3 takes the place of your NH4OH]

3) NH4Cl ==> NH4^+ + Cl^-

Addition of NH4Cl from 3 increases NH4^+ markedly, that makes 2 shift to decrease OH^- which decreases OH^- so much that Ksp for 1 is not exceeded;therefore, the Mg(OH)2 ppt dissolves.

Answer 2

Opps sorry that was a typo, it should be nitric acid. Thanks for the help. I was wondering can you give me any clues for the 2nd experiment, because I quite confused about that as well :/