Answer
The pH of that solution is equal to 1.33
Work Step by Step
1. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 4.7 \times 10^{-2})$
$pH = 1.327902$
2. Determine the right number of significant figures.
The $[H_3O^+]$ had 2 significant figures: $4.7 \times 10^{-2}$.
So, there must be only 2 numbers on the right of the decimal point of the pH.
- Round the number to 2 decimal places:
$pH = 1.33$
