2
$\begingroup$

I wanted to know if there is a way of getting specific quantities of carbonic acid? Carbonic acid is water + $\ce{CO2}$ but not all the $\ce{CO2}$ dissolves in water. So if I wanted $5\:\mathrm{g}$ of carbonic acid solution, how much $\ce{CO2}$ should I add to water? Is there a way to know how much has been added?

$\endgroup$
3
$\begingroup$

To get a specific amount of $\ce{H2CO3}$, you should monitor and adjust the $\mathrm{pH}$ of the water to the $\mathrm{pH}$ that would correspond to $5\:\mathrm{g}$ of $\ce{H2CO3}$. For this you would need to know the acid dissociation constant, $K_\mathrm{a}(\ce{H2CO3})=1.738 \times 10^{-4}$, and the following equations: $$\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$$

$$\ce{H2CO3(aq) +H2O(l) -> HCO3- (aq) +H3O+(aq)}$$

So now we can calculate, using:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]\cdot [\ce{H2O}]}$$

$[\ce{H2O}]$ is omitted, as it remains mostly constant ($55.5084\ \mathrm{M}$). The only time it needs to be considered is when working with extremely concentrated solutions. From this we now get:

$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]}$$

Disregarding any autoprotolysis, the $[\ce{H3O+}]=0$ before any $\ce{CO2}$ is added and any $\ce{H2CO3}$ is formed. Therefore, we know that all $\ce{H3O+}$ comes from the carbonic acid, and that $[\ce{H3O+}]=[\ce{HCO3-}]$. Therefore:

$$[\ce{H2CO3}]=\frac{[\ce{H3O+}]^2}{1.738 \times 10^{-4}}$$

Here, we can greatly simplify this equation, and introduce a parameter that we can actually measure--$\mathrm{pH}$. We know:

$$\mathrm{pH}=-\log[\ce{H3O+}] \Rightarrow [\ce{H3O+}]=10^{-\mathrm{pH}}$$

Substituting, we get:

$$[\ce{H2CO3}]=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}$$

This, however, only gives you the amount of $\ce{H2CO3}$ that is not dissociated into $\ce{H+}$ and $\ce{HCO3-}$. If we want to calculate exactly how much $\ce{H2CO3}$ is in solution, even in the dissociated form, we must add to it the amount that is dissociated, which as you may recall is $[\ce{H3O+}]=10^{-\mathrm{pH}}$:

$$\text{Total}\ \ce{[H2CO3]}=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}+10^{-\mathrm{pH}}$$

From this you can calculate the mass of the $\ce{H2CO3}$ in solution by multiplying this value by the volume of solution present.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Sorry but how will I know that since that will only give me concentration of CO2? would I just say 1 mole of H2CO3 gives 1 mol of CO2? So the concentration of H2CO3 will be the concentration of CO2? $\endgroup$ – user510 Mar 5 '16 at 4:42
  • $\begingroup$ or do you mean to say concentration of CO2 = (10^-pH)^2/(4.3 x 10^-7)? $\endgroup$ – user510 Mar 5 '16 at 4:45
  • $\begingroup$ Yes, sorry. They are equivalent, as by the chemical equation: $\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$. I will clarify. $\endgroup$ – ringo Mar 5 '16 at 5:47
  • $\begingroup$ one more question, although H2O is not included in the equilibrium equation because it is a liquid, wouldn't the quantity of water affect how much CO2 has dissolved? If so, then how do I know how much quantity I should take to make 5g of H2CO3? $\endgroup$ – user510 Mar 5 '16 at 12:53
  • $\begingroup$ This pKa value accounts also reaction of CO2 with water. You need here to use pKa = 3,76 - OP's asking specifically about H2Co3 and it's concentration is much lower then CO2 $\endgroup$ – Mithoron Mar 5 '16 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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