Answer
$pH = 11.694$
Work Step by Step
- Identify the significant electrolytes in $KC_6H_5O$:
$K^+$: Insignificant acid.
$C_6H_5O^-$: Weak base.
Therefore, the only reaction in that solution will be the ionization of this base:
1. Since $C_6H_5O^-$ is the conjugate base of $HC_6H_5O$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.3\times 10^{- 10} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.3\times 10^{- 10}}$
$K_b = 7.692\times 10^{- 5}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HC_6H_5O] = x$
-$[C_6H_5O^-] = [C_6H_5O^-]_{initial} - x = 0.318 - x$
For approximation, we consider: $[C_6H_5O^-] = 0.318M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HC_6H_5O]}{ [C_6H_5O^-]}$
$Kb = 7.692 \times 10^{- 5}= \frac{x * x}{ 0.318}$
$Kb = 7.692 \times 10^{- 5}= \frac{x^2}{ 0.318}$
$ 2.446 \times 10^{- 5} = x^2$
$x = 4.946 \times 10^{- 3}$
Percent ionization: $\frac{ 4.946 \times 10^{- 3}}{ 0.318} \times 100\% = 1.555\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HC_6H_5O] = x = 4.946 \times 10^{- 3}M $
$[C_6H_5O^-] \approx 0.318M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 4.946 \times 10^{- 3})$
$pOH = 2.306$
$pH + pOH = 14$
$pH + 2.306 = 14$
$pH = 11.694$
