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
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.
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 :/
In the first experiment, the reaction involves the reaction between magnesium chloride (MgCl2) and ammonium hydroxide (NH4OH) to form a precipitate. This can be represented by the following equilibrium equation:
MgCl2 (aq) + 2NH4OH (aq) ⇌ Mg(OH)2 (s) + 2NH4Cl (aq)
Initially, when you added the ammonium hydroxide to magnesium chloride, the reaction occurred and resulted in the formation of magnesium hydroxide (Mg(OH)2) as a solid precipitate. This caused the solution to appear cloudy.
However, when you added ammonium chloride (NH4Cl) to the solution, it dissolved the precipitate. This is because ammonium chloride is a source of chloride ions (Cl-) which reacts with the magnesium hydroxide. The equilibrium equation for this dissolution reaction is as follows:
Mg(OH)2 (s) + 2NH4Cl (aq) ⇌ MgCl2 (aq) + 2NH3 (aq) + 2H2O (l)
The ammonium chloride helps to shift the equilibrium towards the dissolution of the precipitate, resulting in a clear and transparent solution.
In the second experiment, the reaction takes place between lead(II) nitrate (Pb(NO3)2) and sodium oxalate (Na2C2O4) to form a precipitate. The equilibrium equation for this reaction can be represented as:
Pb(NO3)2 (aq) + Na2C2O4 (aq) ⇌ PbC2O4 (s) + 2NaNO3 (aq)
When you added sodium oxalate to lead(II) nitrate, it caused the formation of lead(II) oxalate (PbC2O4) as a solid precipitate, resulting in a cloudy solution.
To dissolve the precipitate, you added dilute nitric acid (HNO3), which reacts with the lead(II) oxalate. The equilibrium equation for this dissolution reaction is as follows:
PbC2O4 (s) + 4HNO3 (aq) ⇌ Pb(NO3)2 (aq) + 2CO2 (g) + 2H2O (l)
The nitric acid provides nitrate ions (NO3-) which react with the lead(II) oxalate, helping to dissolve the precipitate. This leads to the solution becoming clearer, although it may still appear slightly cloudy due to the formation of carbon dioxide gas (CO2).
By understanding the equilibrium reactions and the role of the added compounds, you can explain the formation and dissolution of the precipitates in both experiments.