Answer
$pH = 8.87$
Work Step by Step
$Na^+$ has negligible acidity, therefore, we can calculate the pH of a $NaC_2H_3O_2$ solution by calculating the pH of a $C_2H_3{O_2}^-$ solution with the same concentration.
1. Since $C_2H_3{O_2}^-$ is the conjugate base of $HC_2H_3{O_2}$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.556\times 10^{- 10}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [C_2H_3{O_2}^-] = x$
-$[HC_2H_3O_2] = [HC_2H_3O_2]_{initial} - x = 0.1 - x$
For approximation, we consider: $[HC_2H_3O_2] = 0.1M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][C_2H_3{O_2}^-]}{ [HC_2H_3O_2]}$
$Kb = 5.556 \times 10^{- 10}= \frac{x * x}{ 0.1}$
$Kb = 5.556 \times 10^{- 10}= \frac{x^2}{ 0.1}$
$ 5.556 \times 10^{- 11} = x^2$
$x = 7.454 \times 10^{- 6}$
Percent ionization: $\frac{ 7.454 \times 10^{- 6}}{ 0.1} \times 100\% = 0.007454\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [C_2H_3{O_2}^-] = x = 7.454 \times 10^{- 6}M $
$[HC_2H_3O_2] \approx 0.1M$
4. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 7.454 \times 10^{- 6})$
$pOH = 5.13$
$pH + pOH = 14$
$pH + 5.13 = 14$
$pH = 8.87$
