Answer
$[OH^-] = 2.0 \times 10^{- 6}M $
$pH = 8.30$
Work Step by Step
- Since $F^-$ is the conjugate base of $HF$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 3.5\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 3.5\times 10^{- 4}}$
$K_b = 2.9\times 10^{- 11}$
- We have these concentrations at equilibrium:
-$[OH^-] = [HF] = x$
-$[F^-] = [F^-]_{initial} - x = 0.14 - x$
For approximation, we consider: $[F^-] = 0.14M$
- Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HF]}{ [F^-]}$
$Kb = 2.9 \times 10^{- 11}= \frac{x * x}{ 0.14}$
$Kb = 2.9 \times 10^{- 11}= \frac{x^2}{ 0.14}$
$ 4.0 \times 10^{- 12} = x^2$
$x = 2.0 \times 10^{- 6}$
Percent ionization: $\frac{ 2.0 \times 10^{- 6}}{ 0.14} \times 100\% = 0.0014\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HF] = x = 2.0 \times 10^{- 6}M $
$[F^-] \approx 0.14M$
$pOH = -log[OH^-]$
$pOH = -log( 2.0 \times 10^{- 6})$
$pOH = 5.70$
$pH + pOH = 14$
$pH + 5.7 = 14$
$pH = 8.30$
