Answer:
a) pH = 2.573
b) pH = 4.347
Explanation:
a) weak acid: CH3COOH
∴ 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:
∴ 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
The pH of the solution in (a) is 2.57 The pH of the solution in (b) is 4.4.
We have to set up the ICE table for the reaction;
CH3CO2H + H2O ⇄ H3O^+ + CH3CO2^-
I 0.40 0 0
C -x +x +x
E 0.40 - x x x
The pKa of CH3CO2H is 1.8 x 10-5
Hence,
1.8 x 10-5 = x^2/0.40 - x
1.8 x 10-5 (0.40 - x ) = x^2
7.2 x 10-6 - 1.8 x 10-5x = x^2
x^2 + 1.8 x 10^-5x - 7.2 x 10^-6 = 0
x = 0.00267 M
Hence;
pH = -log [0.00267 M] = 2.57
Using the Henderson Hasselbaclch equation;
pH = pKa + log [A-]/[HA]
pKa = - log Ka = -log[1.8 x 10^-5] = 4.7
Hence;
pH = 4.7 + log [0.20 M]/[0.40 M ]
pH = 4.4
Learn more: https://brainly.com/question/5147266
