Answer:
a) pH = 2.88
b) pH = 4.598
c) pH = 5.503
d) pH = 8.788
e) pH = 12.097
Explanation:
∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]
a) 0.0 mL KOH:
mass balance:
⇒ <em>C</em> CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M
charge balance:
⇒ [H3O+] = [CH3COO-]
⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])
⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0
⇒ [H3O+] = 1.314 E.3 M
∴ pH = - Log [H3O+]
⇒ pH = 2.88
b) 5.0 mL KOH:
- CH3COOH + KOH ↔ CH3COONa + H2O
∴ <em>C </em>CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)
⇒ <em>C</em> CH3COOH = 0.05 M
∴ <em>C</em> KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M
charge balance:
⇒ [H3O+] + [K+] = [CH3COO-]
⇒ [CH3COO-] = [H3O+] + 0.033
⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))
⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])
⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]
⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0
⇒ [H3O+] = 2.523 E-5 M
⇒ pH = 4.598
equivalent point:
- (<em>C</em>*V)acid = (<em>C</em>*V)base
⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)
⇒ Vbase = 0.0125L = 12.5 mL
c) 10.0 mL KOH:
∴ <em>C</em> CH3COOH = 0.0143 M
∴ <em>C</em> KOH = 0.057 M
as in the previous point, starting from the mass and charge balances, we obtain:
⇒ [H3O+] = 3.1386 E-6 M
⇒ pH = 5.503
d) 12.5 mL KOH:
at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:
- CH3COO- + H2O ↔ CH3COOH - OH-
∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]
∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M
mass balance:
⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)
charge balance:
⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)
∴ [K+] = <em>C</em> CH3COO- = 0.066 M
(1) = (2):
⇒ [OH-] = [CH3COOH].......(3)
⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])
⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0
⇒ [OH-] = 6.1408 e-6 m
⇒ pOH = 5.212
⇒ pH = 14 - pOH = 8.788
d) 15.0 mL KOH:
after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:
⇒ <em>C</em> KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M
⇒ [OH-] ≅ <em>C</em> KOH = 0.0125 M
⇒ pOH = 1.903
⇒ pH = 12.097
