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I have the equations:

\begin{align} \ce{CO2(g) + H2O(l) &<=> H2CO3(aq)} &\quad K_1 &= 10^{-1.47} \tag{1}\\ \ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)} &\quad K_2 &= 10^{-6.35} \tag{2}\\ \ce{HCO3-(aq) &<=> H+(aq) + CO3^{2-}(aq)} &\quad K_3 &= 10^{-10.33} \tag{3}\\ \end{align}

Atmospheric $\ce{CO2}$ is at $\pu{10^{-3.5} atm}$ and the solution is pure water.

So far I have been able to find the equilibrium equations and have got the concentration of $\ce{H2CO3}$ to be $10^{-4.97}.$ However, from this I do not know where to go in order to find the concentration of bicarbonate ($\ce{HCO3-}$).

I have the equilibrium equations as:

\begin{align} \tag{1} \frac{[\ce{H2CO3}]}{[\ce{CO2}][\ce{H2O}]} &= K_{\ce{CO2}} = 10^{-1.47}\\ \tag{2} \frac{[\ce{H^+}][\ce{HCO3^-}]}{[\ce{H2CO3}]} &= K_1 = 10^{-6.35}\\ \tag{3} \frac{[\ce{H^+}][\ce{CO3^2-}]}{[\ce{HCO3^-}]} &= K_2 = 10^{-10.33} \end{align}

By multiplying $10^{-1.47}$ by $10^{-3.5}$ I got $10^{-4.97}$ as previously stated.

From there I get $\ce{HCO3-}$ to be $10^{-6.35}+\frac{10^{-4.97}}{\ce{H+}}$.

However I do not know how to find $\ce{H+}$.

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    $\begingroup$ Molar concentration of what? It would be ideal if you could provide exact problem and show your calculations. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk ♦
    Mar 12, 2021 at 9:45
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    $\begingroup$ Acidity constant 10^-6.35 already implies ALL CO2(aq) is converted to H2CO3(aq). CO2(aq,tot) + H2O <=> H+ + HCO3-. The truth is, less then 1% of CO2(aq) converts to H2CO3(aq). So it seems to me, by H2CO3(aq) is probably rather consider total CO2(aq) = CO2(aq) + H2CO3(aq) $\endgroup$
    – Poutnik
    Mar 23, 2021 at 11:41
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    $\begingroup$ K(CO2) depends on used units for CO2(g). // the key part is to calculate pH. Relative ratios of carbonic acid / bicarbonate /carbonate depends on that. $\endgroup$
    – Poutnik
    Mar 23, 2021 at 11:47
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    $\begingroup$ Treat the combination of CO2 (aq) and H2CO3 (aq) as any weak acid in pure water, and calculate the pH from the concentration and K1 (you can search on this site or elsewhere for a model calculation). The pH will come out as about 5.7. Then, verify that there will be very little carbonate, and make sure all reactions (including the auto-dissociation of water) are at equilibrium. $\endgroup$
    – Karsten ♦
    Mar 23, 2021 at 15:10
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    $\begingroup$ Please refrain from asking the same question on different platforms, otherwise it might happen that it ends up here twice and quite a few people will be doing the same work also twice. $\endgroup$
    – Martin - マーチン ♦
    Mar 27, 2021 at 0:40

1 Answer 1

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I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$ Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$

$$ \begin{array}{lccc} & \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\ &\text{Initial} & 10^{-4.97} && 0 && 0 \\ &\text{Final} & 10^{-4.97}(1-\alpha) && 10^{-4.97}\alpha && 10^{-4.97}\alpha \end{array} $$

$$K_1 = 10^{-6.35}$$

$$10^{-6.35} = \frac{(10^{-4.97}\alpha)^2}{10^{-4.97}(1-\alpha)}$$

As $K_1 \ll 1,$ $(1 - \alpha)\approx 1:$

$$\alpha = \sqrt{\frac{10^{-6.35}}{10^{-4.97}}}$$

$$ \begin{align} [\ce{HCO3-}] &= \ce{[H+}] = 10^{-4.97}\alpha \\ &= \sqrt{10^{-6.35}\times 10^{-4.97}} \\ &= 10^{-5.66} \approx \pu{2.19E-6 M} \end{align} $$

Note, that I have not considered self ionization of water because the $\mathrm{pH}$ of this solution will be slightly less than $6$ $(\mathrm{pH} = 5.66).$ Which means even after accounting for self ionization of water, the results won't change much. This is a good approximate of what we needed. And if you want to calculate $[\ce{CO3^2-}]$ than just calculate it in a similar way to which you calculated $[\ce{H2CO3}],$ because now you have $K_2,$ $[\ce{HCO3-}]$ and $[\ce{H+}].$

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Nisarg Bhavsar
    Mar 23, 2021 at 17:31
  • $\begingroup$ @NisargBhavsar Please have a look at my edit and feel free to use it as an example. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk ♦
    Mar 23, 2021 at 18:11
  • $\begingroup$ I just edited the and added the table. Thanks, @andselisk for asking to help!! $\endgroup$
    – Nisarg Bhavsar
    Mar 23, 2021 at 18:11
  • $\begingroup$ @NisargBhavsar No prob, the problem with the new Markdown tables is that you have no control over its width and appearance, so it's a poor choice for aligning chemical reactions and ICE tables. $\endgroup$
    – andselisk ♦
    Mar 23, 2021 at 18:12

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