Question

suppose you have just added 200 ml of a solution containing .5000 moles of acetic acid per liter to 100ml of .500 M NaOH. what is the final pH? the Ka of acetic acid is 1.77 x 10^-5

Answer 1

Ka = 1.77 x 10^-5

pKa = 4.75

200 mL of 0.5000 M acetic acid = 0.100 mol of acetic acid.

100 mL of 0.500 M NaOH is 0.05 mol of NaOH.

The reaction between acetic acid and NaOH is as follows:

CH3COOH + NaOH ==> CH3COONa + H2O

0.01 moles NaOH
0.05 moles acetic acid

According to this reaction one mole of acetic acid reacts with one mole of NaOH.

Hence 0.100 mol of acetic acid reacts with 0.100 mol of NaOH.

But here we have 0.05 mol of NaOH, thus 0.05 mole acetic acid left

pH = 4.75 + log 0.05/0.10-0.05

pH = 4.45 + log 0.05/ 0.05

pH = 4.45