Answer
$pH = 8.00$
Work Step by Step
$NaF$: $Na^+$ is not an electrolyte, but $F^-$ acts as a base in water, because it is the conjugate base of a weak acid (HF).
- Since $F^-$ is the conjugate base of $HF$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 7.1\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 7.1\times 10^{- 4}}$
$K_b = 1.4\times 10^{- 11}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$F^-(aq) + H_2O(l) \lt -- \gt HF(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HF$ is : 0 M, and $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HF] = 0 + x = x$
-$[F^-] = [F^-]_{initial} - x $
For approximation, we are going to consider $[F^-]_{initial} = [F^-]$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HF]}{ [F^-]}$
$Kb = 1.4 \times 10^{- 11}= \frac{x * x}{ 0.082}$
$Kb = 1.4 \times 10^{- 11}= \frac{x^2}{ 0.082}$
$ 1.1 \times 10^{- 12} = x^2$
$x = 1.0 \times 10^{- 6}$
Percent ionization: $\frac{ 1.0 \times 10^{- 6}}{ 0.082} \times 100\% = 0.00122\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HF] = x = 1.0 \times 10^{- 6}M $
$[F^-] \approx 0.082M$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 1.0 \times 10^{- 6})$
$pOH = 6.00$
4. Find the pH:
$pH + pOH = 14$
$pH + 6.00 = 14$
$pH = 8.00$
