# What is the equilibrium concentration of "OH"^(-) in a "0.120 M" solution of "Na"_2"CO"_3?

Aug 2, 2017

I got $\text{0.0049 M}$, but you have to be the one who looks up your equilibrium constant to use. That is an important skill...

Well, you have to be the one to look up the ${K}_{a}$ of ${\text{HCO}}_{3}^{-}$, the conjugate acid of carbonate anion. Once you do that, you should then relate back to the ${K}_{b}$, the base association constant, at ${25}^{\circ} \text{C}$ and $\text{1 atm}$:

${K}_{a} {K}_{b} = {K}_{w} = {10}^{- 14}$

$\implies {K}_{b} = {10}^{- 14} / {K}_{a}$

The ${K}_{a}$ is given as $\left[. . .\right]$, and thus, the ${K}_{b}$ is $2.08 \times {10}^{- 4}$. I'll leave it as an exercise for you to figure out the ${K}_{a}$ from your textbook...

Now we simply write the association reaction in water (you'll always do it in water in general chemistry!), and if you wish, you can write an ICE table:

${\text{CO"_3^(2-)(aq) " "+" ""H"_2"O"(l) rightleftharpoons "HCO"_3^(-)(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" ""0.120 M"" "" "" "" "-" "" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "" "-" "" "" "" "+x" "" "" } + x$
$\text{E"" "(0.120 - x)"M"" "-" "" "" "" "x" M"" "" "" "x" M}$

And from this we obtain the equilibrium expression:

$2.08 \times {10}^{- 4} = {x}^{2} / \left(\text{0.120 M} - x\right)$,

with $x$ being the concentration of ${\text{OH}}^{-}$, the stoichiometric decrease in $\left[{\text{CO}}_{3}^{2 -}\right]$ due to conversion to ${\text{HCO}}_{3}^{-}$.

Rearrange to get the quadratic form...

${x}^{2} + 2.08 \times {10}^{- 4} x - \left(2.08 \times {10}^{- 4}\right) \left(0.120\right) = 0$

Solve using the quadratic formula or Wolfram Alpha to get:

color(blue)(ul(x = "0.0049 M" = ["OH"^(-)]))

What is the $\text{pH}$ of this solution?