The precipitate did not dissolve because sulfate ion is too weak a base. The amo
ID: 556249 • Letter: T
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. 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 and magnesium nitrate solution (Ko). Use tabulated equilibrium constants such as K aqueous ammonia K K Kp etc B. To the mixture from Part IV.A., add 10 drops of 6 M acetic acid and mix. ObservationsExplanation / Answer
IV
A) You will observe the formation of white precipitate which is indeed Mg(OH)2
Principal species : Mg+2, NO3-, NH4+, H3O+
2 NH4OH (aq)-----> 2 NH3 (aq)+ H2O (l) ---- (18)
2 NH3 (aq) + H2O (l) + Mg(NO3)2 (aq) -----> Mg(OH)2 (s) + 2 NH4NO3 (aq) --- (19)
2 NH4OH (aq) + Mg(NO3)2 (aq) -----> Mg(OH)2 (s) + 2 NH4NO3 (aq) ---- (20)
B. The precipitate formed in the reaction IV A i.e) Mg(OH)2 will get dissolved in acetic acid.
Mg(OH)2 (s) + 2 CH3COOH (aq) -----> 2 CH3COO- (aq) + Mg2+ + 2H2O (l)
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