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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?

Thank you in advance.

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    $\begingroup$ Pure water is in contradiction with ocean mentioned on the title, as sea water contains 1-4 % of sodium chloride. It affects activity coefficients and dissolved calcium and magnesium ions interact with carbonate. $\endgroup$
    – Poutnik
    Mar 28, 2021 at 14:58
  • $\begingroup$ Thank you @Poutnik, I have changed it now. $\endgroup$
    – Edward
    Mar 28, 2021 at 15:06
  • $\begingroup$ Better than words and numbers is writing applicable algebraic equations. And, as I have recently written somewhere, eq. (2) considers as H2CO3(aq) both H2CO3(aq) and CO2(aq).Their equilibrium constant and both kinetic rate constants are known. $\endgroup$
    – Poutnik
    Mar 28, 2021 at 17:21
  • $\begingroup$ @Poutnik , apologies but I do not appear to be able to see this post you mention. Do you mind providing me with the title? $\endgroup$
    – Edward
    Mar 28, 2021 at 17:55
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    $\begingroup$ Does this answer your question? How to calculate molar concentration from equilibrium equations with no initial concentration given? $\endgroup$
    – Nisarg Bhavsar
    Mar 29, 2021 at 8:08

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