# Question a289d

Apr 7, 2017

Saturated solution of lime has pH value $12.3$. In this solution lime ionizes to form $O {H}^{-}$ ions as follows

$C a {\left(O H\right)}_{2} \left(s\right) \to C {a}^{2 +} \left(a q\right) + 2 O {H}^{-} \left(a q\right)$

When $C {O}_{2} \left(g\right)$ is passed through transparent lime water the acidic $C {O}_{2} \left(g\right)$ first tries to form ${H}_{2} C {O}_{3}$ with water. But ${H}_{2} C {O}_{3}$ formed readily interacts with $O {H}^{-}$ in lime water and forms $C {O}_{3}^{2 -}$ in basic medium i.e at higher pH.

${H}_{2} C {O}_{3} + 2 O {H}^{\text{-}} \to C {O}_{3}^{2 -} + 2 {H}_{2} O$

This $C {O}_{3}^{2 -}$ ions thus formed combine with $C {a}^{2 +}$ ions in solution and produce insoluble $C a C {O}_{3} \left(s\right)$ as white precipitate and the mixture becomes turbid.

$C {a}^{2 +} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right) \to C a C {O}_{3} \left(s\right) \downarrow$

This salt establishes following heterogeneous equilibrium in solution

$C a C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s C {a}^{2 +} \left(a q\right) + C {O}_{3}^{2 -} \left(a q\right) \ldots . . \left[1\right]$

After precipitation of $C a C {O}_{3} \left(s\right)$ is sufficient enough to lower the pH value of the solution considerably then the excess acidic $C {O}_{2} \left(g\right)$ passed through the solution interacts with $C {O}_{3}^{2 -} \left(a q\right)$ ions and water to form $H C {O}_{3}^{-} \left(a q\right)$ as follows

$C {O}_{3}^{2 -} \left(a q\right) + {H}_{2} O + C {O}_{2} \left(g\right) \to 2 H C {O}_{3}^{-} \left(a q\right)$

And thus due to removal of $C {O}_{3}^{2 -} \left(a q\right)$ from solution the equilibrium established before as per equation[1] is shifted towards right and dissolution of $C a C {O}_{3} \left(s\right)$ occurs making the solution transparent again.

Reactions in two steps

$C a {\left(O H\right)}_{2} \left(s\right) \to C a C {O}_{3} \downarrow + {H}_{2} O$

$C a C {O}_{3} + {H}_{2} O + C {O}_{2} \to C a {\left(H C {O}_{3}\right)}_{2}$

Here is how respected Ernest Z likes to modify the answer

The excess carbon dioxide forms hydronium ion, which reacts with the carbonate ions from the calcium carbonate.

Explanation

When $\text{CO"_2"(g)}$ is bubbled through water, it becomes involved in several equilibria:

$\text{CO"_2"(g)" rightleftharpoons "CO"_2"(aq)}$
$\text{CO"_2"(aq)" + "H"_2"O(l)" rightleftharpoons "H"_2"CO"_3"(aq)}$
$\text{H"_2"CO"_3"(aq)" +"H"_2"O(l)" rightleftharpoons "HCO"_3^"-""(aq)" + "H"_3"O"^"+""(aq)}$
$\text{HCO"_3^"-""(aq)" +"H"_2"O(l)" rightleftharpoons "CO"_3^"2-""(aq)" + "H"_3"O"^"+}$

A saturated solution of lime ionizes to form $\text{OH"^"-}$ ions as follows

$\text{Ca(OH)"_2"(s)" rightleftharpoons "Ca"^"2+""(aq)" + "2OH"^"-} \left(a q\right)$

When $\text{CO"_2"(g)}$ is passed through lime water, the $\text{OH"^"-}$ reacts with the $\text{H"_3"O"^"+}$ and pulls the carbon dioxide equilibria to the right.

$\text{CO"_2 + "2OH"^"-" -> "CO"_3^"2-" +"H"_2"O}$

The $\text{CO"_3^"2-}$ ions thus formed combine with $\text{Ca"^"2+}$ ions in solution and produce insoluble $\text{CaCO"_3"(s)}$ as a white precipitate, and the mixture becomes turbid.

$\text{Ca"^"2+""(aq)"+"CO"_3^"2-""(aq)" -> "CaCO"_3"(s)} \downarrow$

The calcium carbonate is involved in the following heterogeneous equilibrium:

$\text{CaCO"_3"(s)"rightleftharpoons"Ca"^"2+""(aq)" + "CO"_3^"2-""(aq)}$

As you continue to bubble in more ${\text{CO}}_{2}$, it forms more $\text{H"_3"O"^"+}$.

The $\text{H"_3"O"^"+}$ reacts with the $\text{CO"_3^"2-}$ from the ${\text{CaCO}}_{3}$.

This pulls the ${\text{CaCO}}_{3}$ equilibrium to the right, and the ${\text{CaCO}}_{3}$ goes back into solution.

The overall reaction is

"CaCO"_3"(s)" rightleftharpoons "Ca"^"2+" + color(red)(cancel(color(black)("CO"_3^"2-")))
$\textcolor{red}{\cancel{\textcolor{b l a c k}{\text{H"_3"O"^"+"))) + color(red)(cancel(color(black)("CO"_3^"2-"))) rightleftharpoons "HCO"_3^"-" + color(red)(cancel(color(black)("H"_2"O}}}}$
"CO"_2 + color(red)(cancel(color(black)("H"_2"O"))) rightleftharpoons color(red)(cancel(color(black)("H"_2"CO"_3")))
$\textcolor{red}{\cancel{\textcolor{b l a c k}{\text{H"_2"CO"_3))) + "H"_2"O" rightleftharpoons "HCO"_3^"-" + color(red)(cancel(color(black)("H"_3"O"^"+}}}}$
stackrel(————————————————————)("CaCO"_3"(s)" + "CO"_2 + "H"_2"O" rightleftharpoons "Ca"^"2+" + "2HCO"_3^"-")#