## General Chemistry: Principles and Modern Applications (10th Edition)

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