What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What percentage of the acid is dissociated?

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO - + Na+

CH3COOH ↔ CH3COO - + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO - ] * [ H3O + ]) / [ CH3COOH ]

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO - ]

∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * (mol/1000mmol) * (1000mL/L)

∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO - ]

charge balance:

⇒ [ H3O + ] + [ Na + ] = [ CH3COO - ] + [ OH - ] ... is negligible [ OH-], comes from water

⇒ [ CH3COO - ] = [ H3O + ] + 1.00

⇒ Ka = (([ H3O + ] + 1) * [ H3O + ]) / (3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O + ]² + [ H3O + ] = 6.12 E-5 - 1.8 E-5 [ H3O + ]

⇒ [ H3O + ]² + [ H3O + ] - 6.12 E-5 = 0

⇒ [ H3O + ] = 6.12 E-5 M

⇒ pH = - Log [ H3O + ] = 4.213

b) (% dis) * mol acid = C CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %