i have come across this question: The mixture formed when 25 cm3 of 2 mol dm–3 sodium hydroxide solution is added to 50 cm3 of 2 mol dm–3 ethanoic acid, for which Ka = 1.7 × 10–5 mol dm–3. (1) A 2.2 B 2.5 C 4.5 D 4.8. And i need to find the PH.
i have got no clue where i have gone wrong my answer was 4.5, but the markscheme says 4.8
i have come across this question: The mixture formed when 25 cm3 of 2 mol dm–3 sodium hydroxide solution is added to 50 cm3 of 2 mol dm–3 ethanoic acid, for which Ka = 1.7 × 10–5 mol dm–3. (1) A 2.2 B 2.5 C 4.5 D 4.8. And i need to find the PH.
i have got no clue where i have gone wrong my answer was 4.5, but the markscheme says 4.8
I believe you have worked out that there are 0.1 mol of EtOOH and 0.05 mol of EtOO-, which you may have converted to concs. then plugged those numbers into the Ka expression and got 4.468521
The correct method would have been to realise that when the 0.05 mol of EtOO- is made, it must have been made out of some of the EtOOH, i.e. the number of mol of EtOOH must go down by the number of mol of EtOO- made (= number of mol of OH- added).
You should have 0.05 mol of both EtOOH AND EtOO- and when you plug the numbers into Ka you'll get 4.769551