I wanted to know if it is possible to add enough carbonic acid to a $\ce{KOH}$ solution to make it an acidic solution.

The $\mathrm{pH}$ of my KOH solution was $13.2$. After adding $\pu{50 g}$ of carbon dioxide (and I do not know how much reacted), the $\mathrm{pH}$ decreased to $10.302$.

Is there a way to know how many moles of carbonic acid more to add to the $\ce{KOH}$ solution to make it acidic?

At first, I thought I could take the change in pH to calculate the additional moles of carbonic acid needed.

So I thought going from a $\mathrm{pH}$ of $10.302$ to $7$, I would calculate the moles of carbonic acid needed for a change in $\mathrm{pH}$ of $3.302$ $\mathrm{pH}$ ($10.302 - 7 = 3.302$). Thus, the calculation would be $10^{-3.302} = \pu{4.989 x 10^{-4} M}$ $\ce{H2CO3}$. There are 2 moles of $\ce{H+}$ in 1 mole of $\ce{H2CO3}$; thus, I need:

$4.989 \times 10^{-4}/2 = 2.49 \times 10^{-4}$ moles of $\ce{H2CO3}$

$\ce{H2CO3}$ is a $1$ to $1$ ratio with $\ce{CO2}$; thus, I need $2.49 \times 10^{-4}$ moles of carbonic acid in solution.

But I do not know if it is right to think it this way?

Can someone tell me if what I am doing is right, and if not, what I can do to theoretically calculate moles of carbonic acid needed to make the $\ce{KOH}$ solution as acidic as I could.

  • $\begingroup$ "as acidic as I could" is too vague. Can you be more specific about, say, your target $\mathrm{pH}$ for the solution? $\endgroup$
    – hBy2Py
    Mar 29, 2017 at 13:48
  • $\begingroup$ between a pH of 5-7 $\endgroup$
    – user510
    Mar 29, 2017 at 13:51
  • $\begingroup$ Are you adding carbonic acid by bubbling pure CO2 through the solution? I think this will be tricky to calculate then assume it will all be absorbed and react (as you found in the first stage of your experiment). Can you add a pH indicator or other means of monitoring pH and just bubble CO2 in via a fine frit until it's neutralized / slightly acidic? I think your calculations represent a minimum of how much you will need in practice. $\endgroup$
    – airhuff
    Mar 29, 2017 at 16:00
  • $\begingroup$ no put dry ice. And @airhuff, you think my calculations are right? $\endgroup$
    – user510
    Mar 29, 2017 at 22:27
  • $\begingroup$ Yea, looks right to me. $\endgroup$
    – airhuff
    Mar 29, 2017 at 22:36

1 Answer 1


You have not stated the volume of your $\ce{KOH}$ solution. I therefore have to make some assumptions.

I will use $\mathrm{p}K_\mathrm{a1} (\ce{H2CO3}) = 6.37$. I further assume you have $\pu{1 L}$ of the $\ce{KOH}$ solution and that this volume will not change by the addition of the carbon dioxide.

For the following, we have also to assume that all the solid carbon dioxide will react, i.e. no evaporation of any gas -- tricky, but perhaps not impossible.

You start with a solution of KOH that has the $\mathrm{pH} = 13.2$. From this we calculate $\mathrm{pOH}$:

$$\mathrm{pOH} = 0.8 \quad \to \quad [\ce{OH-}] = \pu{0.16 M}~(\text{or}~\pu{0.1584 M})$$

First, we suppose that the reaction between the solid carbon dioxide and the solution of KOH will result in only potassium hydrocarbonate:

$$\ce{CO2 + OH- -> HCO3-}$$

OK, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate? You will need (per liter $\ce{KOH}$): $\pu{0.16 mol}$.

After addition of $\pu{0.16 mol}$ carbon dioxide we will face the same situation as if you had dissolved $\pu{0.16 mol}$ of potassium hydrogencarbonate in $\pu{1 L}$ of water. The $\mathrm{pH}$ of such a solution will still be alkaline $(\mathrm{pH} > 8)$.

You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with $\mathrm{pH} = 6.37$.

Accordingly, we have to add more carbon dioxide. The charge balance is:

$$\ce{[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO3^2-]}$$

At $\mathrm{pH} = 6.37$, both $[\ce{OH-}]$ and $[\ce{CO3^2-}] << [\ce{HCO3-}]$. Therefore, we re-write the charge balance as:

$$\ce{[H3O+] + [K+] = [HCO3-]}$$

We know the values of $[\ce{H3O+}]$ and $[\ce{K+}]$, so we calculate:

$$[\ce{HCO3-}] = 10^{-6.37} + 0.16 \approx \pu{0.16 M}$$

We can now calculate $[\ce{H2CO3}]$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} \quad \to \quad [\ce{HCO3-}] = [\ce{H2CO3}] = \pu{0.16 M} \quad \to \quad [\ce{HCO3-}] + [\ce{H2CO3}] = \pu{0.32 M}$$

Accordingly, in total we need $\pu{0.32 mol} \approx \pu{0.3 mol}$ of carbon dioxide to reach $\mathrm{pH} = 6.37$. Per liter KOH solution this will correspond to $\pu{0.3 g} \times \pu{44 g mol-1} = \pu{13 g}$ solid carbon dioxide.

  • $\begingroup$ I reacted 10 grams of KOH in 200mL of water $\endgroup$
    – user510
    Mar 29, 2017 at 23:43
  • $\begingroup$ I am not quite getting the same molar concentration of OH as you are using 13.2 as my pH value. Wouldn't the concentration be 10 raised to the negative pOH? $\endgroup$
    – user510
    Mar 29, 2017 at 23:49
  • $\begingroup$ also, how does CO2 transform hydroxide ions to KHCO3? And I do not understand how you got the charge balance for KHCO3 ([H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO32-]) $\endgroup$
    – user510
    Mar 30, 2017 at 13:53
  • $\begingroup$ It is not the charge balance for KHCO3. It is the charge balance of the system, which tells you that the total concentration of all positive ions in the solution have to be equal to the total concentration of all the negative ions. $\endgroup$
    – Bive
    Mar 30, 2017 at 14:08
  • $\begingroup$ Or better: It is not the charge balance for KHCO3. It is the charge balance of the system, which tells you that the total concentration of positive charges in the solution have to be equal to the total concentration of the negative charges. $\endgroup$
    – Bive
    Mar 30, 2017 at 14:28

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