Question

Thinking about considerations of acid/base chemistry and solubility constants. Calcium carbonate (CaCO₃) is added to initially pure water that is exposed to the atmosphere with 400 ppm(v) of carbon dioxide. If solid CaCO₃ remains after a long time – showing dissolution and precipitation are at equilibrium – what is the aqueous concentration of calcium ions? Use the acid constants, Henry’s Law constants and solubility products

Answer 1

The partial pressure of 400 ppm CO₂ in air, considering atmospheric pressure of the mixture, is 0.3 Torr, or 3.95x10^-4 Atm. Henry's constant for CO₂ in water at 25°C is 29.41 amt/M, so we can calculate the concentration of CO₂ in equilibrium with the air as:

$c=\frac{p}{K_{H}}=\frac{3.95x10^{-4}atm}{29.41atm/M}=1.34x10^{-5}M$

This dissolved CO₂ is in equilibrium with the CO₃^2- ions through two equilibriums:

$CO_{2,(aq)}+H_{2}O_{(l)}\rightarrow HCO_{3,(aq)}^{-}+H^{+}_{(aq)}$

and

$HCO_{3,(aq)}^{-}\rightarrow CO_{3,(aq)}^{2-}+H^{+}_{(aq)}$

With Ka₁ and Ka₂, respectively, which have values of 4.47×10⁻⁷ and 4.69×10⁻¹¹. Using Ka₁ we can calculate the concentration of HCO₃^- and H⁺ (which can be considered the same, since the second dissociation can be considered as negligible when compared with this one):

$Ka_{1}=\frac{[H^{+}][HCO_{3}^{-}]}{[CO_{2}]}=\frac{x^{2}}{[CO_{2}]}$

These concentrations are:

$[H^{+}]=[HCO_{3}^{-}]=\sqrt{[CO_{2}]\cdot Ka_{1}}=\sqrt{1.34x10^{-5}\cdot 4.47x10^{-7}}=2.45x10^{-6}M$

And with this value we can go to the second dissociation (again, we consider the amount of H⁺ generated by this dissociation as negligible:

$Ka_{2}=\frac{[H^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}$

So we get:

$[CO_{3}^{2-}]=\frac{Ka_{2}[HCO_{3}^{-}]}{[H^{+}]}=Ka_{2}=4.69x10^{-11}\: M$

So this is the concentration of carbonate ion coming from the dissolution of CO₂. Since we can consider air as an infinite source of CO₂ at a constant pressure, this concentration is a constant.

With this, we can calculate the amount of dissolved CaCO₃ using its K_sp:

$K_{sp}=1.4x10^{-8}=[Ca^{2+}][CO_{3}^{-}]$

We already have a fixed concentration of CO₃^2- from the dissolution of carbon dioxide, so the rest of the concentration of carbonate will come from the dissolution of CaCO₃, and it will be equal to the concentration of Ca²⁺, since they are both coming from the same compound. If we call this concentrations x, we get:

$1.4x10^{-8}=x\cdot (4.69x10^{-11}+x)=x^{2}+4.69x10^{-11}x$

This is a quadratic equation which can be solved to yield:

$x_{1}=1.18x10^{-4}\: M;x_{2}=-1.18x10^{-4}\: M$

The positive answer is the only one which makes chemical sense, so the desired concentration of Ca²⁺ is 1.18x10^-4 M.