FAD1018 Degree of Dissociation ($\alpha$) — Calculation Guide

Four ways to find $\alpha$. Pick the one that matches what the question gives you.

Recipe 1: Weak Acid/Base from $K$ (Small $\alpha$ Assumption)

Use when they give $K_a$ or $K_b$ and the acid/base is weak.

Steps

# Action Formula
1 Write $K_a = \dfrac{x^2}{c - x}$ ICE table
2 Assume $x \ll c$ so $c - x \approx c$ $K_a \approx \dfrac{x^2}{c}$
3 Solve $x = \sqrt{K_a c}$ $x = [\text{H}_3\text{O}^+]$
4 $\alpha = \dfrac{x}{c}$ $\displaystyle \alpha = \sqrt{\frac{K_a}{c}}$
5 Verify: $%\alpha = \alpha \times 100% < 10%$ If fails → quadratic (Recipe 2)

Worked — Benzoic Acid

0.050 M benzoic acid ($C_6H_5COOH$), $K_a = 6.5 \times 10^{-5}$. Find $\alpha$ and $[H^+]$.

$$x = \sqrt{6.5 \times 10^{-5} \times 0.050} = \sqrt{3.25 \times 10^{-6}} = 1.80 \times 10^{-3}\ \text{M}$$

$$\alpha = \frac{1.80 \times 10^{-3}}{0.050} = 0.0360 \quad (3.60%)$$

Check: $3.60% < 10%$ ✓ Assumption holds.

$$[H^+] = 1.80 \times 10^{-3}\ \text{M}, \quad \text{pH} = 2.74$$

Worked — Methylamine

0.10 M $CH_3NH_2$, $K_b = 4.4 \times 10^{-4}$. Find $\alpha$ and pH.

$$x = \sqrt{4.4 \times 10^{-4} \times 0.10} = \sqrt{4.4 \times 10^{-5}} = 6.63 \times 10^{-3}\ \text{M}$$

$$\alpha = \frac{6.63 \times 10^{-3}}{0.10} = 0.0663 \quad (6.63%)$$

Check: $6.63% < 10%$ ✓ (just barely).

$$[OH^-] = 6.63 \times 10^{-3}\ \text{M}, \quad \text{pOH} = 2.18, \quad \text{pH} = 11.82$$

Recipe 2: Weak Acid/Base from $K$ (Quadratic Required)

Use when Recipe 1's assumption $%\alpha \ge 10%$ fails.

Steps

# Action
1 Set up exact $K_a = \dfrac{x^2}{c - x}$
2 Rearrange: $x^2 + K_a x - K_a c = 0$
3 Solve: $\displaystyle x = \frac{-K_a + \sqrt{K_a^2 + 4K_a c}}{2}$
4 $\alpha = x/c$

Worked — Chloroacetic Acid

0.010 M $ClCH_2COOH$, $K_a = 1.4 \times 10^{-3}$. Check if $\sqrt{K_a/c}$ works.

$$\sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.010}} = \sqrt{0.14} = 0.374 \quad (37.4%)$$

$37.4% \ge 10%$ → assumption fails. Use quadratic.

$$x^2 + (1.4 \times 10^{-3})x - (1.4 \times 10^{-3})(0.010) = 0$$ $$x^2 + 0.0014x - 1.4 \times 10^{-5} = 0$$

$$x = \frac{-0.0014 + \sqrt{(0.0014)^2 + 4(1.4 \times 10^{-5})}}{2} = \frac{-0.0014 + \sqrt{1.96 \times 10^{-6} + 5.6 \times 10^{-5}}}{2}$$

$$x = \frac{-0.0014 + \sqrt{5.796 \times 10^{-5}}}{2} = \frac{-0.0014 + 0.00761}{2} = 0.00311\ \text{M}$$

$$\alpha = \frac{0.00311}{0.010} = 0.311 \quad (31.1%)$$

Compare: the approximate formula gave $37.4%$ — a noticeable error. Always check the assumption.

Recipe 3: Gas-Phase Dissociation via ICE

Use when they give $K_c$ or $K_p$ for a gas-phase equilibrium with changing moles.

Steps (A $\rightleftharpoons$ 2B type)

# Action
1 Set up ICE with initial moles of A, zero B
2 Let $\alpha$ = fraction dissociated; A loses $\alpha$, B gains $2\alpha$
3 Equilibrium moles: A = $1-\alpha$, B = $2\alpha$, total = $1+\alpha$
4 Convert to concentrations (using $[,] = n/V$)
5 Plug into $K_c$ expression, solve for $\alpha$

Worked — Dinitrogen Tetroxide

2.0 mol N$_2$O$_4$ in a 5.0 L flask at a temperature where $K_c = 0.50$. Find $\alpha$ and equilibrium concentrations.

Initial: $[N_2O_4]_0 = 2.0/5.0 = 0.40$ M, $[NO_2]_0 = 0$

N$_2$O$_4$ NO$_2$
I $0.40$ $0$
C $-0.40\alpha$ $+0.80\alpha$
E $0.40(1-\alpha)$ $0.80\alpha$

$$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.80\alpha)^2}{0.40(1-\alpha)} = \frac{0.64\alpha^2}{0.40(1-\alpha)} = \frac{1.6\alpha^2}{1-\alpha}$$

$$0.50 = \frac{1.6\alpha^2}{1-\alpha}$$

$$0.50(1-\alpha) = 1.6\alpha^2$$ $$1.6\alpha^2 + 0.50\alpha - 0.50 = 0$$

$$\alpha = \frac{-0.50 + \sqrt{0.25 + 3.20}}{2(1.6)} = \frac{-0.50 + \sqrt{3.45}}{3.2} = \frac{-0.50 + 1.857}{3.2} = 0.424$$

$$[N_2O_4]{eq} = 0.40(1 - 0.424) = 0.230\ \text{M}$$ $$[NO_2]{eq} = 0.80(0.424) = 0.339\ \text{M}$$

Check: $K_c = (0.339)^2 / 0.230 = 0.50$ ✓

Shortcut (A $\rightleftharpoons$ 2B, from $K_p$ and $P$)

$$\alpha = \sqrt{\frac{K_p}{4P + K_p}} \quad \text{(from } K_p = \frac{4\alpha^2}{1-\alpha^2}P\text{)}$$

Recipe 4: From van't Hoff Factor (Colligative Properties)

Use when the question tells you the observed colligative effect ($\Delta T_f$, $\Delta T_b$, $\Pi$) and the expected value for a non-dissociating solute.

Steps

# Action Formula
1 Calculate $i$ $\displaystyle i = \frac{\text{observed colligative property}}{\text{expected for non-electrolyte}}$
2 Find $n$ = ions per formula unit NaCl $\rightarrow n=2$, CaCl$_2$ $\rightarrow n=3$
3 Solve for $\alpha$ $\displaystyle \alpha = \frac{i - 1}{n - 1}$

Worked — Salt and Freezing Point

0.050 m NaCl has an observed freezing point depression of $0.176\ ^\circ\text{C}$. $K_f = 1.86\ ^\circ\text{C}\cdot\text{m}^{-1}$. Find $\alpha$ of NaCl in this solution.

Expected $\Delta T_f$ for non-electrolyte: $\Delta T_f = K_f m = 1.86 \times 0.050 = 0.093\ ^\circ\text{C}$

$$i = \frac{0.176}{0.093} = 1.892$$

NaCl dissociates into 2 ions ($n = 2$):

$$\alpha = \frac{i - 1}{n - 1} = \frac{1.892 - 1}{2 - 1} = 0.892 \quad (89.2%)$$

Not quite 100% — ion pairing reduces effective dissociation at this concentration.

Recipe 5: From pH (Working Backwards)

Use when they give you the pH of a weak acid/base solution and ask for $\alpha$ directly.

Steps

# Action
1 $[H^+] = 10^{-\text{pH}}$
2 $\displaystyle \alpha = \frac{[H^+]}{c}$
3 Optionally find $K_a$ from $K_a = \dfrac{x^2}{c - x}$

Worked — Formic Acid

A 0.10 M formic acid solution has pH = 2.38. Find $\alpha$ and $K_a$.

$$[H^+] = 10^{-2.38} = 4.17 \times 10^{-3}\ \text{M}$$

$$\alpha = \frac{4.17 \times 10^{-3}}{0.10} = 0.0417 \quad (4.17%)$$

$$K_a = \frac{(4.17 \times 10^{-3})^2}{0.10 - 4.17 \times 10^{-3}} = \frac{1.74 \times 10^{-5}}{0.09583} = 1.82 \times 10^{-4}$$

Quick Pick — Which Recipe?

graph TD
    Q["Question asks for α"] -->|"Weak acid/base + Ka/Kb given"| R1["Recipe 1<br/>α = √(K/c)"]
    R1 -->|"%α ≥ 10%?"| R2["Recipe 2<br/>Quadratic"]
    R1 -->|"%α < 10%?"| D["Done ✓"]
    Q -->|"Gas-phase + Kc/Kp given"| R3["Recipe 3<br/>ICE table"]
    Q -->|"Colligative data + i"| R4["Recipe 4<br/>α = (i-1)/(n-1)"]
    Q -->|"pH given"| R5["Recipe 5<br/>α = [H⁺]/c"]

Common Exam Traps

Trap Fix
Using $\sqrt{K/c}$ without checking $%\alpha$ Always verify $%\alpha < 10%$
Forgetting that $\alpha$ depends on $c$ (unlike $K$) Dilution increases $\alpha$ but $K$ is constant
Confusing $\alpha$ for strong vs weak Strong acids: $\alpha = 1$ always. No $K_a$ calculation needed.
Mixing $K_a$ with $K_b$ For bases: $\alpha = \sqrt{K_b/c}$, then pOH → pH
Wrong $n$ in van't Hoff formula $n$ = ions produced, not atoms in the formula
Gas-phase: using $K_c$ when given $K_p$ Convert: $K_p = K_c(RT)^{\Delta n}$

Exam Shortcuts

  • Small $\alpha$ default: When you see a weak acid with $K_a \le 10^{-4}$ and $c \ge 0.05$ M, the assumption almost always holds.
  • Borderline: If $\alpha \approx 8\text{–}9%$ using the approximation, the quadratic answer will be slightly lower. The approximation is usually close enough unless the question explicitly demands precision.
  • Gas shortcut (A $\rightleftharpoons$ 2B): $\displaystyle \alpha = \sqrt{\frac{K_p}{4P + K_p}}$ — memorise this, it saves the ICE table on MCQ.
  • $\alpha$ from pH: Two steps max. pH → $[H^+]$ → $\alpha = [H^+]/c$.

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