Answer
The pH of that solution is equal to 3.40
Work Step by Step
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[OH^-]$, and solve for $[H_3O^+]$:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.5 \times 10^{- 11} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.5 \times 10^{- 11}}$
$[H_3O^+] = 4.0 \times 10^{- 4}M$
2. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 4.0 \times 10^{- 4})$
$pH = 3.39794$
3. Determine the right number of significant figures.
The $[H_3O^+]$ had 2 significant figures: $4.0 \times 10^{-4}$.
So, there must be only 2 numbers on the right of the decimal point of the pH.
- Round the number to 2 decimal places:
$pH = 3.40$
