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IB DP Chemistry: SL復(fù)習(xí)筆記8.2.5 Acid-Base Calculations

Category: IB課程, 教材筆記, 福利干貨 Date: 2022年8月4日下午4:06

Acid-Base Calculations

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

pH = - log [H⁺]? ??and ???K_w?= [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 10^2?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

8.1.9-Worked-Example-Answer-part-1-2