1. Higher Unit 3 Dilution of acids and alkalis

2. After today’s lesson you should be able to:  Explain what happens to the pH of an acid and an alkali as they are diluted.  Explain what happens to [H + ] and [OH - ] as acids and alkalis are diluted.  Calculate the volume of water required to dilute the pH of an acid or alkali from one pH to another.

3. Dilution of acids and alkalis  As an acid is diluted its pH increases towards 7 as [H + ] decreases.  As an alkali is diluted its pH decreases towards 7 as [OH - ] decreases.

4. Factors of dilution A 1mol l -1 (10 0 mol l -1 ) solution of HCl has a pH of 0. If 10cm 3 of this solution is pipetted into a 100cm 3 volumetric flask and made up to the graduation mark with distilled water then the HCl has been diluted by a factor of 10. The concentration of the diluted solution will be 0.1mol l -1 (10 -1 mol l -1 ) and hence, it will have a pH of 1.

5. To make a solution of pH 2 ([H + ] = 10 -2 mol l -1 ) from a solution of pH 0 ([H + ] = 10 0 mol -1 ) we can pipette 10cm 3 of the pH 0 solution into a 1000cm 3 volumetric flask and make it up to the graduation mark with distilled water i.e. dilute it by a factor of 100.

6. Example 1 How would you make 250cm 3 of a hydrochloric acid solution with pH 2 starting with 100cm 3 of 0.1mol l -1 ? [H + ] = 10 -1 mol l -1 ∴ pH = 1 pH 1 to pH 2 = dilution factor of 10 Need 250cm 3 so pipette 25cm 3 of pH 1 solution into a 250cm 3 volumetric flask and make up to the graduation mark with distilled water.

7. Example 2 How would you make 100cm 3 of a sodium hydroxide solution with pH 11 starting with 250cm 3 of 0.1mol l -1 ? [OH-] = 10 -1 mol l -1 ∴ [H + ] = 10 -14 ÷ 10 -1 = 10 -13 mol l -1 i.e. pH = 13 but require [H + ] = 10 -11 mol l -1 pH 13 to pH 11 = dilution factor of 100 Need 100cm 3 so pipette 1cm 3 of pH 13 solution into a 100cm 3 volumetric flask and make up to the graduation mark with distilled water.

8. Now try……. Questions 13.31 – 13.40 on pages 108 – 109 of ‘Test your Higher Chemistry Calculations’.

9. Practical  Starting with 0.1mol l -1 HCl, how would you make 100cm 3 solution with a pH of 3?  Starting with 0.1mol l -1 NaOH, how would you make 200cm 3 solution with a pH of 12? Now try out both to see if you are correct.