# Question 823c6

Jun 4, 2015

The pH of your solution will be equal to 8.90.

Sodium fluoride, $N a F$, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, $N {a}^{+}$, and fluoride anions, ${F}^{-}$.

$N a {F}_{\left(s\right)} \to N {a}_{\left(a q\right)}^{+} + {F}_{\left(a q\right)}^{-}$

Since 1 mole of sodium fluoride produces 1 mole of fluoride anions, the concentration of the fluoride ions will be equal to that of the salt.

$\left[{F}^{-}\right] = \left[N a F\right] = \text{0.22 M}$

The fluoride ions will react with water to form hydrofluoric acid, a weak acid, and hydroxide ions, $O {H}^{-}$, which is an indicator that the pH of the solution will be greater than 7.

Use an ICE table to determine the concentration of the hydroxide ions

$\text{ } {F}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H {F}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$
I....0.22...............................0..................0
C....(-x)...............................(+x)..............(+x)
E...0.22-x............................x...................x

The base dissociation constant, ${K}_{b}$, will be equal to

${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(3.5 \cdot {10}^{- 5}\right) = 2.86 \cdot {10}^{- 10}$

This means that you'll get

${K}_{b} = \frac{\left[H F\right] \cdot \left[O {H}^{-}\right]}{\left[{F}^{-}\right]} = \frac{x \cdot x}{0.22 - x} = {x}^{2} / \left(0.22 - x\right)$

Since ${K}_{b}$ is so small, you can approximate $\text{(0.22-x)}$ with $0.22$ to get

${K}_{b} = {x}^{2} / 0.22 = 2.86 \cdot {10}^{- 10}$

Therefore,

$x = \sqrt{0.22 \cdot 2.86 \cdot {10}^{- 10}} = 7.9 \cdot {10}^{- 6}$

You can now determine the pOH of the solution

$p O H = - \log \left(\left[O {H}^{-}\right]\right)$

$p O H = - \log \left(7.9 \cdot {10}^{- 6}\right) = 5.10$

As a result, the solution's pH will be

pH_"sol" = 14 - pOH = 14 - 5.10 = color(green)("8.90")#