Salts As Weak Bases Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 × 10^-4.

Question

Salts As Weak Bases
Calculate the pH of a 0.30 M NaF solution. The [latex] K_{a} [/latex] value for HF is 7.2 × [latex] 10^{-4} [/latex].

Accepted Answer

The major species in solution are

[latex] Na^{+}[/latex], [latex] F^{-}[/latex], and [latex]H _{2}O [/latex]

Since HF is a weak acid, the [latex] F^{-}[/latex] ion must have a significant affinity for protons. Therefore the dominant reaction will be

[latex] F^{-}(aq) + H _{2}O (l) \xrightleftharpoons[]{} HF(aq) + OH^{-} (aq)[/latex]

which yields the [latex] K_{b} [/latex] expression

[latex] K_{b} = \frac{[HF] [OH^{-}]}{[F^{-}]} [/latex]

The value of [latex] K_{b} [/latex] can be calculated from [latex] K_{w} [/latex] and the [latex] K_{a} [/latex] value for HF:

[latex] K_{b} = \frac{K_{w}}{ K_{a} (for HF)} =\frac{1.0 × 10 ^{-14}}{7.2 × 10 ^{-4}} = 1.4 × 10 ^{-11} [/latex]

The concentrations are as follows:

Initial Concentration (mol/L)		Equilibrium Concentration (mol/L)
[latex] [F^{-}]_{0} [/latex]=0.30	[latex] \xrightarrow[ H _{2}O to reach equilibrium]{x mol / L F^{-} reacts with } [/latex]	[latex] [F^{-}] [/latex]=0.30 -x
[latex] [HF]_{0} [/latex]=0		[latex] [HF][/latex]= x
[latex] [OH^{-}]_{0} [/latex]≈0		[latex] [OH^{-}][/latex]=x

Thus

[latex] K_{b} = 1.4 × 10 ^{-11} =\frac{[HF] [OH^{-}]}{[F^{-}]} =\frac{(x) (x)}{0.30 - x } ≈ \frac{x²}{0.30} [/latex]

x ≈ [latex] 2.0 × 10 ^{-6} [/latex]

The approximation is valid by the 5% rule, so

[latex] [OH^{-} ]= x= 2.0 × 10 ^{-6} M [/latex]

pOH = 5.69

pH = 14.00 - 5.69 = 8.31

As expected, the solution is basic.

Initial Concentration (mol/L)		Equilibrium Concentration (mol/L)
$[F^{-}]_{0}$ =0.30	$\xrightarrow[ H _{2}O to reach equilibrium]{x mol / L F^{-} reacts with }$	$[F^{-}]$ =0.30 -x
$[HF]_{0}$ =0		$[HF]$ = x
$[OH^{-}]_{0}$ ≈0		$[OH^{-}]$ =x

Question 7.12: Salts As Weak Bases Calculate the pH of a 0.30 M NaF solutio...