# What is the concentration of a solution of K_2CO_3 that has pH = 11.90?

## For ${H}_{2} C {O}_{3}$, ${K}_{a 1} = 4.2 \cdot {10}^{-} 7$, ${K}_{a 2} = 4.8 \cdot {10}^{-} 11$.

Apr 9, 2017

Here's what I got.

#### Explanation:

Potassium carbonate is a soluble ionic compound, which implies that it dissociates completely when dissolved in water to produce potassium cations and carbonate anions.

${\text{K"_ 2"CO"_ (3(aq)) -> 2"K"_ ((aq))^(+) + "CO}}_{3 \left(a q\right)}^{2 -}$

Once in aqueous solution, the carbonate anions will react with water to form bicarbonate anions, ${\text{HCO}}_{3}^{-}$, and hydroxide anions, ${\text{OH}}^{-}$.

"CO"_ (3(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCO"_ (3(aq))^(-) + "OH"_ ((aq))^(-)" "color(darkorange)("(*)")

In essence, the carbonate anion acts as a base in aqueous solution, which is why you have a $\text{pH} > 7$ for this solution.

Now, you know that for

${\text{HCO"_ (3(aq))^(-) + "H"_ 2"O" _ ((l)) rightleftharpoons "CO"_ (3(aq))^(2-) + "H"_ 3"O}}_{\left(a q\right)}^{-}$

you have

${K}_{a 2} = 4.8 \cdot {10}^{- 11}$

As you know, aqueous solutions at room temperature have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{a} \cdot {K}_{b} = {10}^{- 14}}}}$

This means that the base dissociation constant for the carbonate anion will be equal to

${K}_{b} = {10}^{- 14} / \left({K}_{a 2}\right)$

which is

${K}_{b} = \frac{{10}^{- 14}}{4.8 \cdot {10}^{- 11}} = 2.08 \cdot {10}^{- 4}$

Looking at $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, you can write the expression of the base dissociation constant as

${K}_{b} = \left(\left[{\text{HCO"_3^(-)] * ["OH"^(-)])/(["CO}}_{3}^{2 -}\right]\right)$

Now, you can use the $\text{pH}$ of the solution to calculate the equilibrium concentration of hydroxide anions

"pH" = - log(["H"_3"O"^(+)])

hence

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

Aqueous solutions at room temperature have

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}}}}$

You can thus say that the solution has

$\left[{\text{OH}}^{-}\right] = {10}^{- 14} / \left({10}^{- 11.90}\right) = 7.94 \cdot {10}^{- 3}$ $\text{M}$

Once again, looking at $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$, we can say that for every $1$ mole of carbonate anions that ionizes, you get $1$ mole of bicarbonate anions and $1$ mole of hydroxide anions.

This means that the solution will have

$\left[{\text{OH"^(-)] = ["HCO}}_{3}^{-}\right] = 7.94 \cdot {10}^{- 3}$ "M"

You can now determine the equilibrium concentration of the carbonate anions

["CO"_3^(2-)] = (["OH"^(-)] * ["HCO"_3^(-)])/K_b

You will end up with

$\left[{\text{CO}}_{3}^{2 -}\right] = \frac{7.94 \cdot {10}^{- 3} \cdot 7.94 \cdot {10}^{- 3}}{2.08 \cdot {10}^{- 4}}$

["CO"_3^(2-)] = "0.303 M"

This represents the equilibrium concentration of the carbonate anions, i.e. what does not ionize to produce bicarbonate anions and hydroxide anions.

The initial concentration of the carbonate anions must include the concentration that does ionize, so

["CO"_3^(2-)]_0 = "0.303 M" + 7.94 * 10^(-3)"M"

["CO"_3^(2-)]_0 = "0.311 M"

Since potassium carbonate produces carbonate anions in a $1 : 1$ mole ratio, the concentration of the solution will be

color(darkgreen)(ul(color(black)(["K"_2"CO"_3] = "0.31 M")))

The answer is rounded to two sig figs, the number of decimal places you have for the $\text{pH}$ of the solution.