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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

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

Method 1

  1. CH3COOH ↔\leftrightarrow CH3COO- + H+ Ka = 1.74 × 10-5
  2. H2O + H2O ↔\leftrightarrow H3O + OH- Kw = 1.0 ×10-14
    Since Ka >> Kw, : CH3COOH + H2O ↔\leftrightarrow CH3COO- + H3O+
    Ci = 0.05 0 0
    0.05 - 0.5α\alpha 0.05α\alpha 0.05α\alpha
    Ka = (.05a)(.05a)(.05−0.05a)\frac{(.05a)(.05a)}{(.05-0.05a)} = (.05a)(0.05a).05(1−a)\frac{(.05a)(0.05a)}{.05(1-a)} = .05a21−a\frac{.05a^2}{1-a}
    1.74 × 10-5 = .05a21−a\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 = Kac\sqrt{\frac{K_a}{c}}
    a = 1.74×10−5.05\sqrt{\frac{1.74\times10^{-5}}{.05}}
    = 34.8×10−5×1010\sqrt{\frac{34.8\times10^{-5}\times 10}{10}}
    = 3.48×10−6\sqrt{3.48\times 10^{-6}}
    = CH3COOH ↔\leftrightarrow CH3COO- + H+
    α1.86×10−3[CH3COO−]\frac{\alpha 1.86\times10^{-3} }{[CH_3COO^-]}
    = 0.05 × 1.86 × 10-3
    = 0.93×10−31000\frac{0.93 × 10 ^{-3}}{1000}
    = .000093
    ** Method 2**
    Degree of dissociation, a = kac\sqrt{\frac{k_a}{c}}
    c = 0.05 M
    Ka = 1.74 × 10-5
    Then, a = 1.74×10−5.05\sqrt{\frac{1.74\times10^{-5}}{.05}}
    a = 34.8×10−5\sqrt{34.8\times 10^{-5}}
    a = 3.48×10−4\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.