Exercise 4 | Acids and Bases
General Chemistry 3 - Exercise 4
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