IB DP Chemistry: SL复习笔记8.2.5 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
- The pH of a solution of phosphoric acid changes from 3 to 5. Deduce how the hydrogen ion concentration changes
- 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.
- 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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