Question

Determine the pH of (a) a 0.40 M CH3CO2H solution, (b) a solution that is 0.40 M CH3CO2H and 0.20 M NaCH3CO2

Answer 1

Answer:

a) pH = 2.573

b) pH = 4.347

Explanation:

a) weak acid: CH3COOH

• CH3COOH + H2O ↔ CH3COO- + H3O+

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

∴ C CH3COOH = 0.40 M

mass balance:

⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)

charge balance:

⇒ [H3O+] = [CH3COO-].........(2)

(2) in (1):

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )

⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]

⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 2.6743 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.573

b) balanced reations:

• CH3COONa + H2O → Na+  +  CH3COO-
• CH3COOH + H2O ↔ CH3COO-  +  H3O+

∴ C CH3COOH = 0.40 M

∴ C CH3COONa = 0.20 M

mass balanced:

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

⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)

charge balanced:

⇒ [H3O+] + [Na+] = [CH3COO-]

∴ [Na+] = 0.20 M

⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)

(2) in (1):

⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])

⇒ 7.2 E-6  - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]

⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 4.499 E-5 M

⇒ pH = 4.347