Answer
$[OH^-] = 1.24 \times 10^{- 6}M $
$pH = 8.09$
Work Step by Step
- Identify the significant electrolytes in $NaF$:
- $Na^+$ : Insignificant
- $F^-$: Weak base.
Therefore, we just need to calculate the pH of a $0.105M$ $F^-$ solution.
1. Since $F^-$ is the conjugate base of $HF$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 6.8\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 6.8\times 10^{- 4}}$
$K_b = 1.471\times 10^{- 11}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HF] = x$
-$[F^-] = [F^-]_{initial} - x = 0.105 - x$
For approximation, we consider: $[F^-] = 0.105M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HF]}{ [F^-]}$
$Kb = 1.471 \times 10^{- 11}= \frac{x * x}{ 0.105}$
$Kb = 1.471 \times 10^{- 11}= \frac{x^2}{ 0.105}$
$ 1.544 \times 10^{- 12} = x^2$
$x = 1.243 \times 10^{- 6}$
Percent ionization: $\frac{ 1.243 \times 10^{- 6}}{ 0.105} \times 100\% = 0.001183\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HF] = x = 1.243 \times 10^{- 6}M $
$[F^-] \approx 0.105M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.243 \times 10^{- 6})$
$pOH = 5.906$
$pH + pOH = 14$
$pH + 5.906 = 14$
$pH = 8.094$
