How to find the moles of carbonic acid I need to add to make a solution of KOH acidic

Question

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.

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.