# Determine the [OH-] of a solution that is 0.100 M in F-? Determine the pH of this solution?

Oct 4, 2015

$p {H}_{\text{sol}} = 8.07$

#### Explanation:

You actually need the base dissociation constant, ${K}_{b}$, of the fluoride anion, ${\text{F}}^{-}$, in order to be able to calculate the concentration of hydroxide ions it produces in aqueous solution.

Since you did not provide that value, I will use the acid dissociaation constant, ${K}_{a}$, of hydrofluoric acid, $\text{HF}$, to calculate the ${K}_{b}$ of ${\text{F}}^{-}$.

The ${K}_{a}$ of hydrofluoric acid is equal to $7.2 \cdot {10}^{- 4}$. You know that water's self-ionization constant, ${K}_{W}$, is euqal to ${10}^{- 14}$ at room temperature.

Moreover, you know that

${K}_{W} = {K}_{a} \times {K}_{b}$

This means that ${K}_{b}$ will be equal to

${K}_{b} = {K}_{W} / {K}_{a} = {10}^{- 14} / \left(7.2 \cdot {10}^{- 4}\right) = 1.39 \cdot {10}^{- 11}$

So, the fluoride anion will actually react with water to form hydrofluoric acid. Use an ICE table to help you determine the equilibrium concentration of the hydroxide ions

${\text{F"_text((aq])^(-) + "H"_2"O"_text((l]) -> "HF"_text((aq]) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " "0.100" " " " " " " " " " "0" " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " "(+x)" " "(+x)
color(purple)("E")" "(0.100-x)" " " " " "x" " " " " " "x

By definition, the base dissociaation constant will be equal to

${K}_{b} = \left(\left[{\text{HF"] * ["OH"^(-)])/(["F}}^{-}\right]\right) = \frac{x \cdot x}{0.100 - x}$

Because ${K}_{b}$ is so small, you can say that $\left(0.100 - x\right) \approx 0.100$, which means that you have

${K}_{b} = {x}^{2} / 0.100 = 1.39 \cdot {10}^{- 11}$

$x = \sqrt{0.100 \cdot 1.39 \cdot {10}^{- 11}} = 1.18 \cdot {10}^{- 6}$

The equilibrium concentration of hydroxide ions will thus be

x = ["OH"^(-)] = 1.18 * 10^(-6)"M"

The $p O H$ of the solution will be

$p O H = - \log \left(\left[{\text{OH}}^{-}\right]\right) = - \log \left(1.18 \cdot {10}^{- 6}\right) = 5.93$

Therefore, the pH of the solution will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 5.93 = \textcolor{g r e e n}{8.07}$