Exercise 4 | The Properties of Acids and Bases

Calculate the difference in pH values of 2 solutions for which the [H3O+] differ by a factor of 106.

Solution 1:

pH1 = -log [H3O+]1


Solution 2:

pH2 = - log [H3O+]2 with [H3O+]2 = 106 x [H3O+]1


pH2 = - log (106 x [H3O+]1) = - log (106) – log [H3O+]1

⇒ pH2 = – 6 + pH1

The difference is: 6 pH unit