Question

What is the pH when 50 mL 1 M NaOH is added to 450 mL 0.5 M acetic acid?

Answer 1

Intial number of moles of NaOH = 1 * 50 / 1000 = 0.050 mol

Intial moles of acetic acid = 0.5 * 450 / 1000 = 0.225 mol

When NaOH and Acetic acid are combined the following reaction takes place,

NaOH (aq.) + CH3COOH (aq.) ------------------> CH3COONa (aq.) + H2O (l)

So, after the reaction,

Final Moles of acid = 0.225 - 0.050 = 0.175 mol

Final moles of salt formed = 0.050 mol

It is known that the pKa of acetic acid = 4.75

According pH equation of acidic buffers,

pH = pKa + Log[salt]/[acid]

pH = 4.75 + Log(0.050 / 0.175)

pH = 4.206