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Exercises · 6.59

Q.The ionization constant of propanoic acid is 1.32 × 10⁻⁵. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

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For a weak acid, the degree of ionization α\alpha is found from Ka=Cα2/(1−α)K_a = C\alpha^2/(1-\alpha); in pure 0.05 M solution, α≈1.62×10−2\alpha \approx 1.62 \times 10^{-2} and pH ≈ 3.09. In the presence of 0.01 M HCl, the common ion effect suppresses ionization drastically, giving α′≈1.32×10−3\alpha' \approx 1.32 \times 10^{-3}.

Concept First: Weak Acid Ionization

Propanoic acid (CH3CH2COOH\mathrm{CH_3CH_2COOH}) is a weak monoprotic acid. Its ionization in water is an equilibrium:

CH3CH2COOH⇌CH3CH2COO−+H+\mathrm{CH_3CH_2COOH \rightleftharpoons CH_3CH_2COO^- + H^+}

The equilibrium constant Ka=1.32×10−5K_a = 1.32 \times 10^{-5} tells us the acid is weak — only a tiny fraction of molecules actually donate a proton. The degree of ionization α\alpha is the fraction of acid molecules that have ionized. If the initial concentration is CC, then at equilibrium:

  • [HA]=C(1−α)[\mathrm{HA}] = C(1-\alpha)
  • [A−]=Cα[\mathrm{A^-}] = C\alpha
  • [H+]=Cα[\mathrm{H^+}] = C\alpha (from the acid alone)

The key formula is:

Ka=[H+][A−][HA]=(Cα)2C(1−α)=Cα21−αK_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{(C\alpha)^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}

When α\alpha is very small (typically α<0.05\alpha < 0.05), we can approximate 1−α≈11-\alpha \approx 1, giving α≈Ka/C\alpha \approx \sqrt{K_a/C}. But we must check the approximation — if it fails, we solve the quadratic.

1. Pure 0.05 M solution — find α\alpha

Given C=0.05C = 0.05 M, Ka=1.32×10−5K_a = 1.32 \times 10^{-5}.

First, test the approximation: α≈1.32×10−50.05=2.64×10−4=1.62×10−2\alpha \approx \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} = 1.62 \times 10^{-2}.

Is this small enough? 1.62%1.62\% is borderline — the approximation 1−α≈11-\alpha \approx 1 introduces about 1.6%1.6\% error. For exam accuracy (usually 2–3 significant figures), this is acceptable. But let's be thorough and solve exactly.

Write the exact equation:

0.05 α21−α=1.32×10−5\frac{0.05\,\alpha^2}{1-\alpha} = 1.32 \times 10^{-5}

Multiply through:

0.05 α2=1.32×10−5(1−α)0.05\,\alpha^2 = 1.32 \times 10^{-5} (1-\alpha)

0.05 α2=1.32×10−5−1.32×10−5 α0.05\,\alpha^2 = 1.32 \times 10^{-5} - 1.32 \times 10^{-5}\,\alpha

Bring all terms to one side:

0.05 α2+1.32×10−5 α−1.32×10−5=00.05\,\alpha^2 + 1.32 \times 10^{-5}\,\alpha - 1.32 \times 10^{-5} = 0

This is a quadratic in α\alpha. Using the quadratic formula:

α=−1.32×10−5±(1.32×10−5)2+4(0.05)(1.32×10−5)2×0.05\alpha = \frac{-1.32 \times 10^{-5} \pm \sqrt{(1.32 \times 10^{-5})^2 + 4(0.05)(1.32 \times 10^{-5})}}{2 \times 0.05}

The term (1.32×10−5)2=1.74×10−10(1.32 \times 10^{-5})^2 = 1.74 \times 10^{-10} is negligible compared to 4(0.05)(1.32×10−5)=2.64×10−64(0.05)(1.32 \times 10^{-5}) = 2.64 \times 10^{-6}. So:

α≈−1.32×10−5+2.64×10−60.1\alpha \approx \frac{-1.32 \times 10^{-5} + \sqrt{2.64 \times 10^{-6}}}{0.1}

2.64×10−6=1.625×10−3\sqrt{2.64 \times 10^{-6}} = 1.625 \times 10^{-3}

α≈1.625×10−3−1.32×10−50.1=1.6118×10−30.1=1.612×10−2\alpha \approx \frac{1.625 \times 10^{-3} - 1.32 \times 10^{-5}}{0.1} = \frac{1.6118 \times 10^{-3}}{0.1} = 1.612 \times 10^{-2}

Tip

The approximation gave 1.62×10−21.62 \times 10^{-2} — identical to three significant figures. For weak acids where Ka/C<10−3K_a/C < 10^{-3}, the approximation is safe. Here Ka/C=2.64×10−4K_a/C = 2.64 \times 10^{-4}, so it's fine.

So α=1.62×10−2\alpha = 1.62 \times 10^{-2} (or 1.62%1.62\%).

2. pH of the pure solution

[H+]=Cα=0.05×1.62×10−2=8.10×10−4[\mathrm{H^+}] = C\alpha = 0.05 \times 1.62 \times 10^{-2} = 8.10 \times 10^{-4} M.

pH=−log⁡10(8.10×10−4)=3.09\mathrm{pH} = -\log_{10}(8.10 \times 10^{-4}) = 3.09

(Check: log⁡8.1≈0.908\log 8.1 \approx 0.908, so 4−0.908=3.0924 - 0.908 = 3.092.)

3. In presence of 0.01 M HCl — common ion effect

Now the solution already contains H+\mathrm{H^+} from a strong acid (HCl) at 0.01 M. This shifts the weak acid equilibrium to the left — ionization is suppressed.

Let the new degree of ionization be α′\alpha'. The initial concentration of propanoic acid is still C=0.05C = 0.05 M. At equilibrium: …

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