Question

If 10.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.20 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL. Express your answer to two decimal places.

Answer 1

Density of acetic acid = 1.05g/ml

Volume of acetic acid = 10ml

mass of acetic acid = 10ml × 1.05g/ml = 10.5g

Molar mass of CH3COOH = 60.05g/mol

No of mole of CH3COOH = 10.5g/60.05g/mol = 0.17485mol

Volume after dilution = 1.20L

[CH3COOH] = (0.17485mol/1.20L)×1L = 0.1457M

Dissociation of CH3COOH is

CH3COOH(aq) + H2O(l) <------> CH3COO- (aq) + H3O+(aq)

Ka = [CH3COO-] [ H3O+ ]/[CH3COOH] = 1.8 ×10^-5

at equillibrium

[ CH3COO-] = x

[ H3O+ ] = x

[CH3COOH] = 0.1457 - x

Therefore

x^2/(0.1457-x ) = 1.8×10^-5

we can assume 0.1457 - x = 0.1457 because x is small value

x^2/0.1457 = 1.8 × 10^-5

x^2 = 2.62 × 10^-6

x = 1.62 ×10^-3

[ H3O+ ] = 1.62 × 10^-3M

pH = - log(1.62×10^-3)

= 2.79