Answer
$pH = 5.118$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Boric Acid] = [Boric Acid]_{initial} - x = 0.1 - x$
For approximation, we consider: $[Boric Acid] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Boric Acid]}$
$Ka = 5.8 \times 10^{- 10}= \frac{x * x}{ 0.1}$
$Ka = 5.8 \times 10^{- 10}= \frac{x^2}{ 0.1}$
$ 5.8 \times 10^{- 11} = x^2$
$x = 7.616 \times 10^{- 6}$
Percent ionization: $\frac{ 7.616 \times 10^{- 6}}{ 0.1} \times 100\% = 0.007616\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [Conj. Base] = x = 7.616 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[Boric Acid] \approx 0.1M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 7.616 \times 10^{- 6})$
$pH = 5.118$
