Question 46: The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
1) CH3COOH ↔ CH3COO- + H+ Ka = 1.74x10-5
2) H2O + H2O ↔ H3O+ + OH- Kw = 1.0x10-14
Since Ka > Kw
CH3COOH + H2O ↔ CH3COO- + H3O+
Ci = 0.05 0 0
0.05- 0.05α 0.05α 0.05α
ka = 0.05α x 0.05α / 0.05- 0.05α
= 0.05α x 0.05α / 0.05 (1-α)
= 0.05α2 / (1-α)
⇒ 1.74x10-5 = 0.05α2 / (1-α)
⇒ 1.74x10-5 - 1.74x10-5 α = 0.05α2
⇒ 0.05α2 + 1.74x10-5 α - 1.74x10-5 =0
D = b2 - 4ac
= (1.74x10-5 )2 - 4 (0.05) (1.74x10-5)
= 3.02x10-25 + 0.348x10-5

Method 2:
Degree of Dissociation,

CH3COOH ↔ CH3COO- + H+
Thus, concentration of CH3COO- = c.α
= 0.05x1.86x10-2
= 0.93x10-2
=.00093M
Since [oAc-] = [H+]
[H+] = .00093 = 0.093x10-2
pH = -log[H+]
= -log (0.93x10-2)
∴ pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.
Comments
Excellent advice