Question

Calculate the pH of a 0.20 M acetic acid solution after adding 0.05M sodium hyroxide (Ka= 1.78*10^-5 M) [Hint: You can use two separate ICE tables (acetic acid + hydroixde, then acetic acid + water), or the Henderson-Hasselbac hequation could be used.]

Answer 1

Find out the concentration of acetic acid and acetate ion.

Lets assume volume = 1.0 L

Find out the moles of acetic acid and NaOH

Moles of acetic acid = 0.20 M x 1.0 L = 0.20 mol

Moles of NaOH = 0.05 M x 1.0 L = 0.05 mol

Reaction

CH3COOH + NaOH --- > CH3COONa + H2O

Mol ratio between CH3COOH and NaOH is 1:1

Moles of NaOH are less than CH3COOH therefore it is reacted completely and 0.05 mol CH3COONa is formed.

Moles of CH3COOH = original moles – moles of NaOH = 0.20 – 0.05 = 0.15 mol CH3COOH

Since volume is 1.0 , moles of both = molarity

[CH3COOH ] = 0.15 M

[CH3COONa] = [CH3COO- ] = 0.05 M

Lets use Henderson-hasselbalch equation.

pH = pka + log ([CH3COO-]/[CH3COOH]

pka = - log ka = - log 1.78E-5 = 4.7496

pH = 4.7496 + log ( 0.05 M / 0.15 M)

= 4.27

pH of the solution would be 4.27