It's been a long time since I did my chemistry classes and I'm currently trying to analyze groundwater samples for hydrogeology purposes. I would like to evaluate carbonate and bicarbonate concentration from groundwater samples, but I only have values of total alkalinity as $\ce{CaCO3}$, $\mathrm{pH}$, and temperature. Is it possible? Some of the $\mathrm{pH}$ values are above 8.3.

I know that:

  • If $\mathrm{pH}$ is above 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+2[\ce{CO3^2-}]+[\ce{OH-}]-[\ce{H+}]$
  • If $\mathrm{pH}$ is under 8.3: $[\mathrm{alk}_{tot}]=[\ce{HCO3-}]+[\ce{OH-}]-[\ce{H+}]$

With the $\mathrm{pH}$, I can find calculate $[\ce{OH-}]$ and $[\ce{H+}]$. But how can I calculate $[\ce{HCO3-}]$ and $[\ce{CO3^2-}]$?

Thank you!

  • $\begingroup$ I remember getting 2 values, for titration to phenolphthaleinum ( if alkalic enough ) and methyl orange titration ends. The first was took for carbonates only and MO for carbonate + bicarbonate weighed sum. But it is my memory for chemical high school, focused on analytical chemistry in 1980-84 and subsequest undergrad lectures and labs. $\endgroup$ – Poutnik Apr 21 at 6:47
  • $\begingroup$ Remember that Henderson-Hasselbalch provides the equilibrium ratio of concentrations at a given pH. $\endgroup$ – Zhe Apr 23 at 16:45

If I understood your question correctly, you have solutions where you know there is a given amount of calcium carbonate dissolved, and would like to know the distribution of this carbonate between all the species present. This proportion is commonly refered as the alpha($\alpha$) for a given species, that varies from 0 to 1(0% - 100%). From your question, I can make some assumptions:

  • As a groundwater sample, any solids dissolved are very diluted, so we don't need to worry about ionic strength;
  • pH is not fixed;
  • Temperature is not fixed, but I will assume its close to room temperature;
  • As other components are not mentioned, I will assume all carbonate comes from calcium carbonate.

Carbonic acid, $\ce{H2CO3}$, has two ionizable hydrogens, so it may assume three forms: The free acid itself, bicarbonate ion, $\ce{HCO3-}$(first-stage ionized form) and carbonate ion $\ce{CO3^2+}$(second-stage ionized form). The respective proportions in comparison with the total concentration of calcium carbonate dissolved are $\alpha0$, $\alpha1$ and $\alpha2$. They must sum to 1(100%), as in chemical reactions matter is neither created or destroyed, only changing between forms. When the calcium carbonate dissolves, a equilibrium is established between its three forms, expressed by the respective equilibrium equations:

First stage: $$\ce{H2O + H2CO3 <=> H3O+ + HCO3-}$$ $$K1 = \frac{\ce{[H3O+][HCO3-]}}{\ce{[H2CO3]}} \approx 4.47*10^-7 $$

Second stage: $$\ce{H2O + HCO3- <=> H3O+ + CO3^2-}$$ $$K2 = \frac{\ce{[H3O+][CO3^2-]}}{\ce{[HCO3-]}} \approx 4.69*10^-11 $$

You can also write a equation for the overrall reaction, by sum of each stage (and multiplication of the respective equilibrium constants): $$\ce{2H2O + H2CO3 <=> 2H3O+ + CO3^2-}$$ $$K1K2 = \frac{\ce{[H3O+]^2[CO3^2-]}}{\ce{[H2CO3]}}$$

Analysing our system, to give a full treatment, if we know the solution pH, we can calculate $\ce{[H3O+]}$. So we are left with three unknown variables, $\ce{[H2CO3]}$, $\ce{[HCO3-]}$ and $\ce{[CO3^2+]}$. But so far we have only two independent mathematical equations, for K1 and K2 (the overrall equation does't count as independent, as it's only the merging together of the other two). To solve it, we need at least one more independent equation, to match the number of unknows. What we need is the equation for the material balance of the system. As we assumed all carbonate came from calcium carbonate, we can write: $$Cs = \ce{[CaCO3]} = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$

Where Cs here stands for the known concentration of the salt, calcium carbonate. Now we can start replacing values taken from the equilibrium expressions into the material balance, isolating each unknow. For the bicarbonate, for example: $$Cs = \ce{[H2CO3] + [HCO3-] + [CO3^2-]}$$ $$Cs = \ce{\frac{[HCO3-][H3O+]}{K1} + [HCO3-] + \frac{K2[HCO3-]}{[H3O+]}}$$ $$Cs = \ce{\frac{[HCO3-][H3O+]^2 + K1[HCO3-][H3O+] + K1K2[HCO3-]}{K1[H3O+]}}$$ $$\frac{\ce{[HCO3-]}}{Cs} = \ce{\frac{K1[H3O+]}{[H3O+]^2 + K1[H3O+] + K1K2}} = \alpha1$$

So we got the expression for $\alpha1$, that has a curious structure: a fraction, where the denominator is a polynomial of degree 2, and the numerator its middle term. The same procedure can be repeated to find the expressions for the alphas of the other dissolved species. For sake of brevity, I won't do it, but the final result will be: $$\alpha0 = \frac{\ce{[H2CO3]}}{Cs} = \ce{\frac{[H3O+]^2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$ $$\alpha2 = \frac{\ce{[CO3^2-]}}{Cs} = \ce{\frac{K1K2}{[H3O+]^2 + K1[H3O+] + K1K2}}$$

Note that a interesting pattern emerges. The expressions for the remaining two species have the same structure, just changing the term that goes in the numerator.

With the expressions for all species, it's helpful to use a spreadsheet to automate the calculations for a entire range of pH values, to grasp in a visual way what happens with carbonates as pH changes. I did just that, look at the results (here the spreadsheet, to whomever wants to download and play with it): enter image description here

We see that in lower pH the predominant form for carbonate is the free carbonic acid. A bit over 6 bicarbonate ion takes over, and reigns up to pH a bit over 10, from where fully ionized carbonate ion takes over. For a given pH, the concentration of each species can be computed multiplying the respective $\alpha$ by the concentration of total calcium carbonate originally present.

The plot that looks like a "XX" also allows us to see a interesting property of carbonates. Nowhere in the plot you will find a pH value where we have the three species all in significant amounts. In the lower pH region you can find both bicarbonate and carbonic acid. But carbonate only shows up when carbonic acid goes away. It's like the unconfortable situation where you have two close friends who both hate each other. In a given moment I can see you in a room talking with either friend, but I will never see you three in the same room, or both friends of yours. The dividing line is close to the pH 8.6 you mentioned in your question. However, that sad situation has a upside. It makes the problem easier to calculate.

The full treatment I gave to this problem was indeed overkill. If I have three species, but only two show up together at any given time, I can "forget" I'm dealing with a diprotic acid. I need only to see the dividing line I've found, around pH 8.6. If I'm above it, free carbonic acid concentration is zero, and I have to deal only with the pair bicarbonate/carbonate, pretending the bicarbonate anion is just a monoprotic acid. From the equilibrium, we have: $$\ce{[H3O+]} = \frac{\ce{K2[HCO3-]}}{\ce{[CO3^2-]}}$$

Or in logarithimic form: $$pH = pK2 + log(\frac{\ce{[HCO3-]}}{[CO3^2-]})$$

This is the old Henderson–Hasselbalch equation you surely heard about before. As we know the pH and K2, we can calculate the ratio between carbonate and bicarbonate.

In the other side, if I'm below my dividing line near 8.6, carbonate ion concentration is zero, now I have to deal only with the pair carbonic acid/bicarbonate, pretending carbonic acid is just other monoprotic acid. From the equilibrium, we have: $$\ce{[H3O+]} = \frac{\ce{K1[H2CO3]}}{\ce{[HCO3-]}}$$

Or in logarithimic form: $$pH = pK1 + log(\frac{\ce{[H2CO3]}}{[HCO3-]})$$

As we know the pH and K1, we can calculate the ratio between carbonic acid and bicarbonate.

If you want to study in depth such calculations, I recommend this book: Butler, James N. Ionic Equilibrium: Solubility and PH Calculations. John Wiley & Sons, 1998.

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  • $\begingroup$ Thank you so much! What if the temperature is lower than or higher than room temperature? Does it change the "K" values? $\endgroup$ – Rachel Apr 23 at 14:06
  • $\begingroup$ Yes, they do. If a exact result is desired, it's necessary to account for that, and use the constants corrected for the actual temperature. But unless the difference in temperature is big, the error will be probably acceptable. Look this question: chemistry.stackexchange.com/questions/9108/… $\endgroup$ – ksousa Apr 23 at 15:54

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