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]
∴ <em>C</em> 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+
∴ <em>C</em> CH3COOH = 0.40 M
∴ <em>C</em> CH3COONa = 0.20 M
mass balanced:
⇒ <em>C</em> CH3COOH + <em>C</em> 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
