Question

The precipitate did not dissolve because sulfate ion is too weak a base. The amount of sulfate ion that reacts with the hydronium ion is too small to cause a shift in the equilibrium sufficient to dissolve a signiticant amount of the insoluble sulfate. In general, the anion of a precipitate must be a stronger base than fluoride ion in order for the precipitate to be dissolved by the addition of a strong acid. After the reactions with barium ion, rinse the mixtures from the well plate into your waste beaker. Do not dilute this mixture. Pour the contents of the waste beaker into the waste bottle in the fume hood. IV. Other equilibrium reactions. A. Dispense 10 drops of 6 M ammonia. Add 10 drops of 0.1 M magnesium nitrate solution. Save the mixture. Observations: Principal species: a. b. Write two equations that could be thought of as steps in this reaction. Add the two equations together to give the overall equation, Eq. 20. (18) (19) (20) Determine the value of the equilibrium constant for the overall reaction between aqueous ammonia and magnesium nitrate solution (K). Use tabulated equilibrium constants such as KnK, K, etc. B. To the mixture from Part IV.A., add 10 drops of 6 M acetic acid and mix. Observations

Answer 1

IV. A
A formation of white precipitate called Mg(OH)₂ can be observed from this reaction
Major Principal species present will be Mg⁺², NO₃^-, NH₄⁺, H₃O⁺

You will observe that aqueous ammonia will present as stated in the equation below,
2 NH₄OH (aq)-----> 2 NH₃ (aq)+ H₂O (l) ------ (18)
and the next reaction will be,
2 NH₃ (aq) + H₂O (l) + Mg(NO₃)₂ (aq) -----> Mg(OH)₂ (s) + 2 NH₄NO₃ (aq) --- (19)
Add (18) and (19) to get (20)
2 NH₄OH (aq) + Mg(NO₃)₂ (aq) -----> Mg(OH)₂ (s) + 2 NH₄NO₃ (aq) ---- (20)

Tabulated values are not given to solve this problem

B. The precipitate formed in the reaction IV A i.e) Mg(OH)₂ will get dissolved in acetic acid
and therefore the final equation will look like,
Mg(OH)₂ (s) + 2 CH₃COOH (aq) -----> 2 CH₃COO^- (aq) + Mg²⁺ + 2H₂O (l)