Acid-Base Equilibria

Question

How many moles og H30+ or OH- must you add to a liter of strong-acid solution to adjust its pH from 3.15 to 3.65? Assume a negligible volume change?

Please help me with this one, a youtube video will be much appreciated :)

Answer 1

pH = 3.15; [H₃o⁺] = 10^-3.15 = 7.08 x 10^-4 M or mol/L

pH = 3.65; [H₃o⁺] = 10^-3.65 = 2.24 x 10^-4 M

To adjust the pH of the strong acid solution from 3.15 to 3.65, one needs to remove

7.08 x 10^-4 M - 2.24 x 10^-4 M = 4.84 x 10^-4 M H₃O⁺

To remove/neutralize H₃O⁺, we need to add OH^- like the following equation

H₃O⁺ + OH^- -> 2H₂O

Moles of OH^- needed

4.84 x 10^-4 M H₃O⁺ x (1 mol OH^- / 1 mol H₃O⁺) = 4.84 x 10^-4 M OH^- ____ Answer!