Saturated solution of lime has pH value #12.3#. In this solution lime ionizes to form #OH^-# ions as follows
#Ca(OH)_2(s)->Ca^(2+)(aq) + 2OH^(-)(aq)#
When #CO_2 (g)# is passed through transparent lime water the acidic #CO_2 (g)# first tries to form #H_2CO_3# with water. But #H_2CO_3# formed readily interacts with #OH^-# in lime water and forms #CO_3^(2-)# in basic medium i.e at higher pH.
#H_2CO_3+2OH^"-" ->CO_3^(2-)+2H_2O#
This #CO_3^(2-)# ions thus formed combine with #Ca^(2+) # ions in solution and produce insoluble #CaCO_3(s)# as white precipitate and the mixture becomes turbid.
#Ca^(2+)(aq)+CO_3^(2-)(aq)-> CaCO_3(s) darr#
This salt establishes following heterogeneous equilibrium in solution
#CaCO_3(s)rightleftharpoonsCa^(2+)(aq)+CO_3^(2-)(aq).....[1]#
After precipitation of #CaCO_3(s)# is sufficient enough to lower the pH value of the solution considerably then the excess acidic #CO_2 (g)# passed through the solution interacts with #CO_3^(2-)(aq)# ions and water to form #HCO_3^(-)(aq)# as follows
#CO_3^(2-)(aq)+H_2O+CO_2 (g)->2HCO_3^(-)(aq)#
And thus due to removal of #CO_3^(2-)(aq)# from solution the equilibrium established before as per equation[1] is shifted towards right and dissolution of #CaCO_3(s)# occurs making the solution transparent again.
Reactions in two steps
#Ca(OH)_2(s)->CaCO_3darr+H_2O#
#CaCO_3+H_2O+CO_2->Ca(HCO_3)_2#
Please note
Here is how respected Ernest Z likes to modify the answer
Answer:
The excess carbon dioxide forms hydronium ion, which reacts with the carbonate ions from the calcium carbonate.
Explanation
When #"CO"_2"(g)"# is bubbled through water, it becomes involved in several equilibria:
#"CO"_2"(g)" rightleftharpoons "CO"_2"(aq)"#
#"CO"_2"(aq)" + "H"_2"O(l)" rightleftharpoons "H"_2"CO"_3"(aq)"#
#"H"_2"CO"_3"(aq)" +"H"_2"O(l)" rightleftharpoons "HCO"_3^"-""(aq)" + "H"_3"O"^"+""(aq)"#
#"HCO"_3^"-""(aq)" +"H"_2"O(l)" rightleftharpoons "CO"_3^"2-""(aq)" + "H"_3"O"^"+"#
A saturated solution of lime ionizes to form #"OH"^"-"# ions as follows
#"Ca(OH)"_2"(s)" rightleftharpoons "Ca"^"2+""(aq)" + "2OH"^"-"(aq)#
When #"CO"_2"(g)"# is passed through lime water, the #"OH"^"-"# reacts with the #"H"_3"O"^"+""# and pulls the carbon dioxide equilibria to the right.
#"CO"_2 + "2OH"^"-" -> "CO"_3^"2-" +"H"_2"O"#
The #"CO"_3^"2-"# ions thus formed combine with #"Ca"^"2+" # ions in solution and produce insoluble #"CaCO"_3"(s)"# as a white precipitate, and the mixture becomes turbid.
#"Ca"^"2+""(aq)"+"CO"_3^"2-""(aq)" -> "CaCO"_3"(s)"darr#
The calcium carbonate is involved in the following heterogeneous equilibrium:
#"CaCO"_3"(s)"rightleftharpoons"Ca"^"2+""(aq)" + "CO"_3^"2-""(aq)"#
As you continue to bubble in more #"CO"_2#, it forms more #"H"_3"O"^"+"#.
The #"H"_3"O"^"+"# reacts with the #"CO"_3^"2-"# from the #"CaCO"_3#.
This pulls the #"CaCO"_3# equilibrium to the right, and the #"CaCO"_3# goes back into solution.
The overall reaction is
#"CaCO"_3"(s)" rightleftharpoons "Ca"^"2+" + color(red)(cancel(color(black)("CO"_3^"2-")))#
#color(red)(cancel(color(black)("H"_3"O"^"+"))) + color(red)(cancel(color(black)("CO"_3^"2-"))) rightleftharpoons "HCO"_3^"-" + color(red)(cancel(color(black)("H"_2"O")))#
#"CO"_2 + color(red)(cancel(color(black)("H"_2"O"))) rightleftharpoons color(red)(cancel(color(black)("H"_2"CO"_3")))#
#color(red)(cancel(color(black)("H"_2"CO"_3))) + "H"_2"O" rightleftharpoons "HCO"_3^"-" + color(red)(cancel(color(black)("H"_3"O"^"+")))#
#stackrel(————————————————————)("CaCO"_3"(s)" + "CO"_2 + "H"_2"O" rightleftharpoons "Ca"^"2+" + "2HCO"_3^"-")#
