# Physical Chemistry: Specifically the Dissolution of Carbonates

Please note this question is different to the other that keeps being mentioned. That one is about the concentration of just $$\ce{HCO3-}$$ when no $$\ce {pH}$$ was given. Whereas, in this one I am trying to find the concentrations of $$\ce{H+}$$, $$\ce{HCO3-}$$, $$\ce{CO3^{2-}}$$, and $$\ce{Ca^2+}$$, with a given $$\ce {pH}$$.

Consider the following equations:

\begin{align} \ce{CO2(g) + H2O(l) &<=> H2CO3(aq)} &\quad K_{\ce{CO2}} &= 10^{-1.47} \tag{1}\\ \ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)} &\quad K_1 &= 10^{-6.35} \tag{2}\\ \ce{HCO3-(aq) &<=> H+(aq) + CO3^{2-}(aq)} &\quad K_2 &= 10^{-10.33} \tag{3}\\ \ce{CaCO3(s) &<=> Ca^{2+}(aq) + CO3^{2-}(aq)} &\quad K_{\text{cal}} &= 10^{-8.48} \tag{4}\\ \end{align}

Atmospheric $$\ce{CO2}$$ is at $$\pu{10^{-3.5} atm}$$ and the solution is pure water. Also, in equation $$(4)$$ calcite crystals are added in excess ($$\ce{CaCO3}$$).

So far I have been able to find the equilibrium equations and have got the concentration of $$\ce{H2CO3}$$ to be $$10^{-4.97}$$ by multiplying the atmospheric $$\ce{CO2}$$ by the value of $$K_{\ce{CO2}}$$.

Having been told the $$\mathrm{pH}$$ of the solution is 8.26, I then calculated the concentration of $$\ce{H+}$$ to be $$10^{-8.26}$$.

Using my values of $$\ce{H2CO3}$$ and $$\ce{H+}$$ and substituting them into the equilibrium equation for (2),I have the concentration of $$\ce{HCO3-}$$ to be $$10^{-3.06}$$. Is this correct?

Continuing on with the same method I have the following concentrations for $$\ce{CO3^2-}$$ and $$\ce{Ca^2+}$$ to be $$10^{-5.13}$$ and $$10^{-3.35}$$, respectively. Am I still correct here?