Question

Consider the carbonate ion test that you might perform on a typical +2 metal carbonate as an equilibrium:

MCO3 + 2H3O+ 3H2O = M(H2O)6^2+ +CO2

A) based on this equilibrium and the observed (limited) solubility of CO2 in water, predict the outcome of bubbling CO2 gas through solutions of each of them three +1/+2 cations. Justifly tour prediction.

B) In light of your answer to part (a) of this question, explain why a solution of Zn(H2O)2(OH)4^2- upon treatment with CO2 gas, first gives a precipitate of ZNCO3; then the precipitate dissolves on continued addition of CO2. Write equations for both reactions involved.

Answer 1

a) For +1 cation like Na⁺ bubbling with CO₂ gives bicarbonate as below

Na₂CO₃ (saturated) + H₂O + CO₂ = 2 NaHCO₃↓ .

For +1 cation like Li⁺ bubbling with CO₂ gives bicarbonate as below

Li₂CO₃ (solid) + H₂O + CO₂ → 2LiHCO₃ (solution)

For +1 cation like K⁺ bubbling with CO₂ gives bicarbonate as below

K₂CO₃ (solid) + H₂O + CO₂ → 2KHCO₃ (solution)

For +2 cation like Ca²⁺bubbling with CO₂ gives bicarbonate as below

CaCO₃ (s) + H₂O (l) + CO₂ (aq) → Ca(HCO₃)₂ (aq)

b) [Zn(H₂O)₂(OH)₄]^2- upon treatment with CO₂ gas, first gives a precipitate of ZnCO₃

[Zn(H₂O)₂(OH)₄]^2- (aq) + CO₂ (g) → ZnCO₃ (s) + 3H₂O (l) + 2OH^- (aq)

On continued addition of CO₂ we get bicarbonate precipitate, wherein bicarbonate being unstable dissolves in solution.

ZnCO₃(s) + CO₂(g) + H₂O(l) → Zn(HCO₃)₂ (aq) (soluble)