IB Chemistry SL

Revision Notes

8.1.9 Acid-Base Calculations

Acid-Base Calculations

  • Using the relationships between pH, [H] and [OH] a variety of problems can be solved

pH = – log [H+]    and    Kw = [H+] [OH]

  • Test your understanding on the following worked examples:

Worked Example

  1. The pH of a solution of phosphoric acid changes from 3 to 5. Deduce how the hydrogen ion concentration changes
  2. Water from a pond was analysed and found to have a hydrogen ion concentration of 2.6 x 10-5 mol dm-3. Calculate the pH of the pond water.
  3. Determine the pH of a solution made by dissolving 5.00 g of potassium hydroxide in 250 cm3 of distilled water

Answers:

Answer 1: The initial pH of the phosphoric acid is 3 which corresponds to a hydrogen ion concentration of 1 x 10-3 mol dm-3 :

[H+] = 10-pH

[H+] = 1 x 10-3 mol dm-3

 The final pH is 5, which corresponds to 1 x 10-5 mol dm-3

Therefore, the solution has decreased in [H+] concentration by 102 or 100 times

 

Answer 2: The pond water has [H+] = 2.6 x 10-5 mol dm-3.

pH = – log [H+] =   -log(2.6 x 10-5) = 4.58

 

Answer 3: Potassium hydroxide (M = 56.10 g mol-1) is a strong base so the concentration of [OH] is the same as the concentration of the solution as it fully dissociates:

KOH (s)      K+ (aq) + OH(aq)

The concentration of KOH is

Using Kw = [H+][OH], and then rearranging  [H+] = Kw /[OH]

pH = – log (2.80 x 10-14) = 13.55

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