Answer
2.981 L of water.
Work Step by Step
Equations that will be used:
$[H^+] = 10^{-pH}$
$C_1 * V_1 = C_2 * V_2$
1. Find the $[H^+]$ that we have:
$[H^+] = 10^{-2.50} = 3.162 \times 10^{-3} M$
2. Find the $[H^+]$ that we want:
$[H^+] = 10^{-3.10} = 7.943 \times 10^{-4}M$
3. Knowing that we have 1L of solution, find with how many "L" we will have of that concentration
$3.162 \times 10^{-3} * 1 = 7.943 \times 10^{-4} * V_2$
$\frac{3.162 \times 10^{-3}}{7.943 \times 10^{-4}} = V_2$
$V_2 = 3.981 L$
We will need the same solution with 3.981 L of water, therefore we need to add 2.981* L to get that pH.
* 3.981 L - 1L = 2.981L
