Question

What is the concentration of a solution of ${K_2}C{O_3}$ that has pH = 11.9$$$$?

Answer 1

First we have to find the concentration of each element of the given compound. Then to calculate the concentration of an aqueous solution you need to know the concentration of the hydronium ion in moles per litre(molarity). The concentration is then calculated using the expression: pH = - log [H3O+].

Complete step by step answer:
Potassium carbonate is a soluble ionic compound, which implies that it dissociates completely when dissolved in water to produce potassium cations and carbonate anions.
${K_2}C{O_{3(aq)}} \to \to 2K_{(aq)}^ + + CO_{3(aq)}^{2 - }$
Once in aqueous solution, the carbonate anions will react with water to form bicarbonate anions, $HCO_3^ -$, and hydroxide anions,$O{H^ - }$.
$CO_{3(aq)}^{2 - } + {H_2}{O_{(I)}} \rightleftharpoons HCO_{3(aq)}^ - + OH_{(aq)}^ -$--(1)
In essence, the carbonate anion acts as a base in aqueous solution, which is why you have a $pH > 7$ for this solution.
Now, you know that for
$HCO_{3(aq)}^ - + {H_2}{O_{(I)}} \rightleftharpoons CO_{3(aq)}^{2 - } + {H_3}O_{(aq)}^ -$
you have ${K_{a2}} = 4.8 \times {10^{ - 11}}$
As you know, aqueous solutions at room temperature have
${K_a} \times {K_b} = {10^{ - 14}}$
This means that the base dissociation constant for the carbonate anion will be equal to
${K_b} = \dfrac{{{{10}^{ - 14}}}}{{{K_{a2}}}}$ which is ${K_b} = \dfrac{{{{10}^{ - 14}}}}{{4.8 \times {{10}^{ - 11}}}} = 2.08 \times {10^{ - 4}}$
Looking at the equation (1), you can write the expression of the base dissociation constant as
Now, you can use the pH of the solution to calculate the equilibrium concentration of hydroxide anions
$pH = - \log \left( {\left[ {{H_3}{O^ + }} \right]} \right)$ hence $\left[ {{H_3}{O^ + }} \right] = {10^{ - pH}}$
Aqueous solutions at room temperature have
You can thus say that the solution has
$\left[ {O{H^ - }} \right] = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 11.90}}}} = 7.94 \times {10^{ - 3}}M$
Once again, looking at the equation (1), we can say that for every one mole of carbonate anions that ionizes, you get one mole of bicarbonate anions and one mole of hydroxide anions.
This means that the solution will have
$\left[ {O{H^ - }} \right] = \left[ {HCO_3^ - } \right] = 7.94 \times {10^{ - 3}}M$
You can now determine the equilibrium concentration of the carbonate anions
You will end up with
$\left[ {CO_3^{2 - }} \right] = \dfrac{{7.94 \times {{10}^{ - 3}} \times 7.94 \times {{10}^{ - 3}}}}{{2.08 \times {{10}^{ - 4}}}} = 0.303M$
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
$\left[ {CO_3^{2 - }} \right] = 0.303M + 7.94 \times {10^{ - 3}} = 0.311M$
Since potassium carbonate produces carbonate anions in a $1:1$ mole ratio, the concentration of the solution will be
$\left[ {{K_2}C{O_3}} \right] = 0.31M$

Note:
Note that there is another method to find the concentration of the molecule when mass and volume is given. Divide the mass of the solute by the total volume of the solution. Write out the equation $C = \dfrac{m}{V}$, where $m$ is the mass of the solute and $V$ is the total volume of the solution. Plug in the values you found for the mass and volume, and divide them to find the concentration of your solution.