0
$\begingroup$

I have a problem, in which the concentration of $\ce{Ca^{2+}}$ ions in water should be related to the partial pressure of $\ce{CO2}$ (assuming the water is an open system and in equilibrium with atmosphere).

All the approaches I saw so far, started by looking at the charge ballance of this system:

$$\ce{2 [Ca^{2+}] + [H+]} =\ce{2[CO3^{2-}] + [HCO3-] + [HO-]}$$

This is totally logical. But they continue with the assumption that the terms $\ce{[CO3^{2-}]}$,$\ce{[HO-]}$, $\ce{[H+]}$ are insignificantly small so that you can work with:

$$\ce{2 [Ca^{2+}]} =\ce{[HCO3-]}$$

That's the point I don't get.

Assuming at the beginning there is only $\ce{CaCO3}$ and Water. Both dissociate and let's say $\ce{CO3^{2-}}$ immediatly combines with $\ce{H+}$ to $\ce{HCO3-}$, then it is reasonable that the contribution of $\ce{H+}$ and $\ce{CO3^{2-}}$ can be ignored. But what about the remaining $\ce{OH-}$? Stupidly written, I would get something like:

$$\ce{H2O + CaCO3 } =\ce{ HCO3- + Ca^{2+} + OH-}$$

So why can I also neglect the charge caused by $\ce{OH-}$ Ions? For each bicarbonate/calcium-ion produced I would also get a hydronium ion. why can this still be ignored?

Improve this question
$\endgroup$
3
  • $\begingroup$ Is this pure water to start with, or something like ocean water that is buffered by other species already in solution? In other words, is the pH determined by $\ce{CaCO3}$ and $\ce{CO2}$ or set to some value? $\endgroup$
    – Karsten ♦
    Commented Mar 23, 2019 at 15:55
  • $\begingroup$ Please state the problem that you're trying to solve exactly as it was given to you. Pointless to be guessing as to what the problem actually is. $\endgroup$
    – MaxW
    Commented Mar 23, 2019 at 17:57
  • $\begingroup$ @Karsten Theis : He mentioned water in equilibrium with the aerial CO2, therefore CaCO3 dissolution with CO2 support, forming carbonate hardness Ca(HCO3)2 $\endgroup$
    – Poutnik
    Commented Mar 23, 2019 at 19:44

2 Answers 2

Reset to default
3
$\begingroup$

This is not totally unreasonable.

Calcium carbonate as solubility product of $K_{\mathrm{sp}} = 3.3\times 10^{-9}$ (source).

This means that if you're dissolving the solid, you'd expect a maximum calcium ion concentration of $\ce{[Ca^{2+}]} = 5.7\times 10^{-5}\ \mathrm{M}$. This is a 2.5 orders of magnitude larger than either hydronium or hydroxide concentration in neutral solution. You'd expect the carbonic acid to be mostly dissociated (at least 50%), so the bicarbonate concentration will be similar to the calcium ion concentration (maybe factor of 2 smaller).

The moment you step away from a neutral solution though, this analysis totally falls apart. You'd need more details on the assumptions of the model in that case.

Improve this answer
$\endgroup$
8
  • $\begingroup$ My official Android client says at 2 places "Undefined control sequence \pu" (While otherwise displaying chemical formatting correctly) $\endgroup$
    – Poutnik
    Commented Mar 23, 2019 at 15:20
  • $\begingroup$ @orthocresol I think \pu is valid for us, but may not render correctly in general. $\endgroup$
    – Zhe
    Commented Mar 23, 2019 at 19:38
  • $\begingroup$ @Poutnik Official Android client is great for receiving notifications and reading Stack Overflow. Other Stack Exchange websites are poorly supported (e.g. tables and arrays will also produce the error) and I would recommend to use your web browser instead. See Undefined control sequence \pu in official SE Android app; MathJax in the Android App for more technical details. $\endgroup$
    – andselisk ♦
    Commented Mar 24, 2019 at 8:55
  • 1
    $\begingroup$ I see. I will have it on mind. Otherwise, the client is very convenient for reading. Well written. $\endgroup$
    – Poutnik
    Commented Mar 24, 2019 at 8:57
  • 1
    $\begingroup$ @andselisk Well, it looks fine on browsers now, after your change, but it destroyed rendering in the client, that was originally OK. :-). But as you say, that was expected. Thanks for update. The new version seems more difficult to edit. I use Android 9. $\endgroup$
    – Poutnik
    Commented Mar 24, 2019 at 12:57
1
$\begingroup$

In short, $[\ce{H+}]$, $[\ce{OH-}]$ and $[\ce{CO3^2-}]$ are kept low by ion recombination, honouring the dissociation constants of water and involved compounds, mentioned below.

The reaction mentioned in the question $$\ce{H2O + CaCO3 v -> HCO3−+Ca^2+ + OH−}$$ can be reformulated as $$\ce{CO3^2- + H2O <=> HCO3- + OH- } $$

As there is abundance of $\ce{H+}$, created by the dissoved aerial $\ce{CO2}$, $\ce{OH-}$ and $\ce{CO3^2-}$ quickly recombine with other ions.

$$\begin{align} \ce{OH- + H+ &<=>> H2O} \\ \ce{OH- + HCO3- &<=>> CO3^2 + H2O} \\ \ce{CO3^2- + H+ &<=>> HCO3-} \\ \ce{HCO3- + H+ &<=> H2CO3 <=> CO2 + H2O} \\ \end{align}$$

Note that pure water saturated by the aerial $\ce{CO2}$ has $\mathrm{p}H=5.65$. At the beginning of $\ce{CaCO3}$ dissolution, $\ce{[OH−]} $ is not allowed to be > $\pu{4.10^-9 mol / l)}$.

The aerial $\ce{CO2}$ presence in water shifts the dissolution equilibrium. the calcium concentration is several orders higher than it would be in the pure, carbon dixode free water.

The major reversible summary equation below is basis of the calcite and aragonite cave phenomena:

$$\ce{CO2 + H2O + CaCO3 <=> Ca^2+ + 2 HCO3^-}$$

Typical $c_\ce{Ca^2+}$ and $c_\ce{HCO3-}$ in natural water are in order of units of $\pu{mmol/l}$.
$c_\ce{Ca^2+} = \pu{1 mmol/l}$ is equivalent to $\pu{5.6 dGH}$ (dGH on Wikipedia).

$[\ce{H+}]$ and $[\ce{OH-}]$ are limited by the water dissociation constant.

$$[\ce{H+}][\ce{OH-}] = K_\mathrm{w} = 10^{-14} \quad(\text{at}~\pu{25 °C})$$

If we consider water $\mathrm{pH}$ in $6-8$ range, $[\ce{H+}]$ and $[\ce{OH-}]$ are below $\pu{1 μmol/l}$, over 1000 times less than $[\ce{HCO3-}]$.

For $\mathrm{pH} = 7$

$$\begin{align} \frac{[\ce{CO3^2-}]}{[\ce{HCO3-}]} &= 0.000469 \\ \frac{[\ce{HCO3-}]}{[\ce{CO2}]} &= 4.25 \end{align}$$

So again, $\ce{[CO3^2-]}$ is in range of $\pu{1 μmol/l}$

The $\ce{CO2}$ related equilibrium constants from Wikipedia:

$$ \begin{align} \frac{[\ce{H+}][\ce{CO3^2-}]}{[\ce{HCO3-}]} &= K_\mathrm{a2} = \pu{4.69e-11} \\ \frac{[\ce{H+}][\ce{HCO3-}]}{[\ce{H2CO3}]} &= K_\mathrm{a1} = \pu{2.5e-4} \\ \frac{[\ce{H2CO3}]}{[\ce{CO2}]} &= K_\mathrm{h} = \pu{1.7e-3} \end{align} $$

Improve this answer
$\endgroup$
5
  • $\begingroup$ I think downvoting without a comment or an edit suggestion is not much constructive. Note that the equilibrium concentration of Ca+2 and HCO3- is much higher than the other responder noted,, as CO2 presence shifts the dissolution equilibrium. $\endgroup$
    – Poutnik
    Commented Mar 23, 2019 at 14:58
  • $\begingroup$ I will comment without downvoting: This answer is hard to read because you are not typesetting the chemical equations. The argument is difficult to understand because it is more of a data collection. What would be helpful would be the concentration of all species in a saturated solution of $\ce{CaCO3}$ that has a vapor pressure of $\ce{CO2}$ equal to the partial pressure present in the atmosphere today. $\endgroup$
    – Karsten ♦
    Commented Mar 23, 2019 at 15:48
  • $\begingroup$ Thanks for the comment. It is interesting it could be hard to read,but perhaps I am too used to ASCII based Usenet of the old days of the Internet. I will try to improve it. $\endgroup$
    – Poutnik
    Commented Mar 23, 2019 at 15:51
  • $\begingroup$ @andselisk While learning, I have noticed I should not use \ce for variable subscripts, because of italic convention. It is a pity, as \ce is more compact than that mathxx string. :-) $\endgroup$
    – Poutnik
    Commented Mar 24, 2019 at 13:32
  • $\begingroup$ @Poutnik \ce{...} is for chemical formulas only (you can remember it mnemonically as chemical expression), using it otherwise breaks readability and may cause weird (aka unexpected) MathJax behavior. As for italics, have a look at the quick guide Which symbols are written in roman (upright) font and which are italicized?. I made many mistakes at the beginning, but after numerous corrections from the Mod team I finally got used to it:) $\endgroup$
    – andselisk ♦
    Commented Mar 24, 2019 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.