Answer
$pH = 4.85$
Work Step by Step
$CH_3COONa$: $Na^+$ is not an electrolyte, but $CH_3COO^-$ acts as a base:
- Since $CH_3COO^-$ is the conjugate base of $CH_3COOH$ , we can calculate its $K_b$ by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.6\times 10^{- 10}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$CH_3COO^-(aq) + H_2O(l) \lt -- \gt CH_3COOH(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $CH_3COOH$ is : 0 M, and $[OH^-]$ $\approx$ 0 M
-$[OH^-] = [CH_3COOH] = 0 + x = x$
-$[CH_3COO^-] = [CH_3COO^-]_{initial} - x $
For approximation, we are going to consider $[CH_3COO^-]_{initial} = [CH_3COO^-]$
2. Now, use the $K_b$ value and equation to find the 'x' value.
$Kb = \frac{[OH^-][CH_3COOH]}{ [CH_3COO^-]}$
$Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.36}$
$Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.36}$
$ 2.0 \times 10^{- 10} = x^2$
$x = 1.4 \times 10^{- 5}$
Percent ionization: $\frac{ 1.4 \times 10^{- 5}}{ 0.36} \times 100\% = 0.0039\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3COOH] = x = 1.4 \times 10^{- 5}M $
$[CH_3COO^-] \approx 0.36M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.4 \times 10^{- 5})$
$pH = 4.85$
