Answer
$pH = 8.52$
Work Step by Step
- $K^+$ has insignificant acidity, so, we just have to consider the $CH{O_2}^-$ ionization:
1. Since $CH{O_2}^-$ is the conjugate base of $HCHO_2$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 4}}$
$K_b = 5.556\times 10^{- 11}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [HCHO_2] = x$
-$[CH{O_2}^-] = [CH{O_2}^-]_{initial} - x = 0.2 - x$
For approximation, we consider: $[CH{O_2}^-] = 0.2M$
3. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HCHO_2]}{ [CH{O_2}^-]}$
$Kb = 5.556 \times 10^{- 11}= \frac{x * x}{ 0.2}$
$Kb = 5.556 \times 10^{- 11}= \frac{x^2}{ 0.2}$
$ 1.111 \times 10^{- 11} = x^2$
$x = 3.333 \times 10^{- 6}$
Percent ionization: $\frac{ 3.333 \times 10^{- 6}}{ 0.2} \times 100\% = 0.001667\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HCHO_2] = x = 3.333 \times 10^{- 6}M $
$[CH{O_2}^-] \approx 0.2M$
4. Calculate the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 3.333 \times 10^{- 6})$
$pOH = 5.48$
$pH + pOH = 14$
$pH + 5.48 = 14$
$pH = 8.52$
