Answer
The pH of this solution is equal to $9.0$.
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for $[H_3O^+]$:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1 \times 10^{- 5} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 5}}$
$[H_3O^+] = 1 \times 10^{- 9}M$
2. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1 \times 10^{-9})$
$pH = 9.000000$
3. Determine the right number of significant figures.
The $[H_3O^+]$ had 1 significant figure: $1 \times 10^{-9}$.
So, there must be only one number on the right of the decimal point of the pH.
- Round the number to only one decimal place:
$pH = 9.0$
