Question

The ionization constant of acetic acid is 1.74 × 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

Method 1
1) CH₃COOH \(\leftrightarrow\) CH₃COO^- + H⁺               K_a = 1.74 × 10^-5
2) H2O + H2O \(\leftrightarrow\) H3O + OH^-               K_w = 1.0 ×10^-14
Since K_a >> K_w, : CH₃COOH + H₂O \(\leftrightarrow\) CH₃COO^- + H₃O⁺
C_i =                         0.05                             0               0 
                            0.05 - 0.5\(\alpha\)                  0.05\(\alpha\)         0.05\(\alpha\)
K_a = \(\frac{(.05a)(.05a)}{(.05-0.05a)}\) = \(\frac{(.05a)(0.05a)}{.05(1-a)}\) = \(\frac{.05a^2}{1-a}\)
1.74 × 10^-5 = \(\frac{.05a^2}{1-a}\)= 1.74 × 10^-5 - 1.74 × 10^-5a = 0.05a² = 0.05a² + 1.74 × 10^-5
D = b² - 4ac = (1.74 × 10^-5)² - 4(0.5)(1.74 × 10^-5) = 3.02 × 10^-25 + .348 × 10^-5
a = \(\sqrt{\frac{K_a}{c}}\)
a = \(\sqrt{\frac{1.74\times10^{-5}}{.05}}\) 
= \(\sqrt{\frac{34.8\times10^{-5}\times 10}{10}}\) 
= \(\sqrt{3.48\times 10^{-6}}\) 
= CH₃COOH \(\leftrightarrow\) CH₃COO^- + H⁺
\(\frac{\alpha 1.86\times10^{-3} }{[CH_3COO^-]}\) 
= 0.05 × 1.86 × 10^-3 
= \(\frac{0.93 × 10 ^{-3}}{1000}\) 
= .000093
Method 2
Degree of dissociation, a = \(\sqrt{\frac{k_a}{c}}\)
c = 0.05 M
K_a = 1.74 × 10^-5
Then, a = \(\sqrt{\frac{1.74\times10^{-5}}{.05}}\)
a = \(\sqrt{34.8\times 10^{-5}}\)
a = \(\sqrt{3.48\times10^{-4}}\)
a = 1.8610^-2
CH₃COOH \(\leftrightarrow\) CH₃COO^- + H⁺
Thus, concentration of CH3COOa^-= c.a = 0.5 × 1.86 × 10^-2 = .093 × 10^-2 = .093 × 10^-2 = .000093 M
Since [oAc^-] = [H⁺],
[H⁺] = .00093 = .093 ×10^-2.
pH = - log[H⁺] = - log(.093 × 10^-2)
∴ pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.