Exercises — Strong vs weak acids - bases; degree of dissociation α

What we use: by definition, $\alpha$ is the dissociated fraction, and each dissociated HA gives exactly one H⁺, so $[{\text{H}}^{+}]=c\alpha$.

$\alpha =\frac{[{\text{H}}^{+}]}{c}=\frac{2.0\times 10^{-3}}{0.20}=1.0\times 10^{-2}$

As a percentage: $\alpha =1.0\%$.

Interpretation: out of every 100 HA molecules, only 1 has split into H⁺ + A⁻. The other 99 sit undissociated. That is what "weak" means numerically.

HCl is a strong acid → it dissociates essentially completely → $\alpha \approx 1$. CH₃COOH is a weak acid → only a small fraction splits → $\alpha \ll 1$.

Larger α → more H⁺ → lower pH. So Beaker A (HCl) has both the larger α and the lower pH. Same label on the bottle ($0.10$ M) does not mean same acidity — only α × concentration decides how many H⁺ are actually free.

Step 1 — is the approximation allowed? Check the ratio $c/K_a$: $\frac{c}{K_a}=\frac{0.050}{1.3\times 10^{-5}}=3846\gg 100✓$ Since this is far above 100, α will be under 5%, so $1-\alpha \approx 1$ is safe.

Step 2 — apply $\alpha \approx \sqrt{K_a/c}$: $\alpha \approx \sqrt{\frac{1.3\times 10^{-5}}{0.050}}=\sqrt{2.6\times 10^{-4}}=1.612\times 10^{-2}=1.61\%$

Step 3 — pH: $[{\text{H}}^{+}]=c\alpha =0.050\times 1.612\times 10^{-2}=8.06\times 10^{-4}\text{M}$ $\text{pH}=-{\log }_{10}(8.06\times 10^{-4})=3.09$

Symmetry with acids: using the $K_b$ and $K_w$ definitions recapped at the top of this page, a weak base $\text{B}+{\text{H}}_2\text{O}\rightleftharpoons {\text{BH}}^{+}+{\text{OH}}^{-}$ obeys the same algebra with $K_b$ in place of $K_a$ and OH⁻ in place of H⁺.

$\frac{c}{K_b}=\frac{0.10}{1.8\times 10^{-5}}=5556\gg 100✓$ $\alpha \approx \sqrt{\frac{K_b}{c}}=\sqrt{\frac{1.8\times 10^{-5}}{0.10}}=\sqrt{1.8\times 10^{-4}}=1.342\times 10^{-2}=1.34\%$ $[{\text{OH}}^{-}]=c\alpha =0.10\times 1.342\times 10^{-2}=1.342\times 10^{-3}\text{M}$ $\text{pOH}=-{\log }_{10}(1.342\times 10^{-3})=2.87$ (and pH would be $14-2.87=11.13$, using $\text{pH}+\text{pOH}=14$ from $K_w$.)

(a) Approximate: ${\alpha }_{\text{approx}}=\sqrt{\frac{1.0\times 10^{-3}}{1.0\times 10^{-2}}}=\sqrt{0.10}=0.316=31.6\%$ But $c/K_a=10$, which is $\ll 100$ — a red flag that this is unreliable.

(b) Exact quadratic $\alpha =\frac{-K_a+\sqrt{K_a^2+4c{K}_a}}{2c}$ (derived at the top of this page): $K_a^2=1.0\times 10^{-6},4c{K}_a=4(10^{-2})(10^{-3})=4.0\times 10^{-5}$ $\sqrt{1.0\times 10^{-6}+4.0\times 10^{-5}}=\sqrt{4.1\times 10^{-5}}=6.40\times 10^{-3}$ ${\alpha }_{\text{exact}}=\frac{-1.0\times 10^{-3}+6.40\times 10^{-3}}{2(10^{-2})}=\frac{5.40\times 10^{-3}}{2.0\times 10^{-2}}=0.270=27.0\%$

(c) Relative error: $\frac{0.316-0.270}{0.270}=0.171=17.1\%\text{too high.}$ Look at figure s02: this is exactly the region (left end, small $c$) where the blue (approx) curve peels above the orange (exact) curve.

${\alpha }_{0.10}=\sqrt{\frac{1.8\times 10^{-5}}{0.10}}=1.342\times 10^{-2}=1.34\%$ ${\alpha }_{0.0010}=\sqrt{\frac{1.8\times 10^{-5}}{0.0010}}=\sqrt{1.8\times 10^{-2}}=0.1342=13.4\%$

Factor change in α: $0.1342/0.01342=10$. So a 100× dilution raised α by exactly $\sqrt{100}=10$ — because $\alpha \propto 1/\sqrt{c}$ (Ostwald's law).

What about [H⁺]? $[{\text{H}}^{+}{]}_{0.10}=0.10\times 0.01342=1.342\times 10^{-3}\text{M}$ $[{\text{H}}^{+}{]}_{0.0010}=0.0010\times 0.1342=1.342\times 10^{-4}\text{M}$ $[{\text{H}}^{+}]$ falls by 10×, even though α rose 10×. Reason: $[{\text{H}}^{+}]=c\alpha \propto c\cdot c^{-1/2}=c^{1/2}$, so it tracks $\sqrt{c}$ downward. This is exactly what figure s01 shows: the blue α curve climbs to the left while the orange $[{\text{H}}^{+}]$ curve falls to the left. Dilution frees a larger fraction but of a much smaller total.

(a) $[{\text{H}}^{+}]=10^{-\text{pH}}=10^{-2.80}=1.585\times 10^{-3}$ M.

(b) $\alpha =\frac{[{\text{H}}^{+}]}{c}=\frac{1.585\times 10^{-3}}{0.25}=6.34\times 10^{-3}=0.634\%$.

(c) Feed straight into the exact expression: $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.25(6.34\times 10^{-3}{)}^2}{1-6.34\times 10^{-3}}=\frac{0.25\times 4.02\times 10^{-5}}{0.9937}=1.01\times 10^{-5}$ Because α is tiny, $1-\alpha \approx 1$, so the approximate $K_a=c{\alpha }^2=1.005\times 10^{-5}$ agrees to three digits. This acid is comparable in strength to acetic acid.

Setup. With both acid and its conjugate base at $0.10$ M we have a buffer. The acetate we poured in dwarfs the tiny amount produced by dissociation, so we treat $[\text{HA}]\approx 0.10$ and $[{\text{A}}^{-}]\approx 0.10$.

$K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}\Rightarrow [{\text{H}}^{+}]=K_a\frac{[\text{HA}]}{[{\text{A}}^{-}]}=1.8\times 10^{-5}\times \frac{0.10}{0.10}=1.8\times 10^{-5}\text{M}$

New α of the acid = extra H⁺ produced per acid ÷ acid concentration: $\alpha =\frac{[{\text{H}}^{+}]}{c}=\frac{1.8\times 10^{-5}}{0.10}=1.8\times 10^{-4}=0.018\%$

Compare: α crashed from $1.34\%$ to $0.018\%$ — about 74× smaller. The flood of acetate ions pushes the equilibrium $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$ hard to the left (Le Chatelier), suppressing dissociation. That suppression is exactly what makes a buffer resist pH change.

(a) Naïve: $\alpha \approx \sqrt{K_a/c}=\sqrt{1.8\times 10^{-5}/1.0\times 10^{-4}}=\sqrt{0.18}=0.424$. But $c/K_a=5.56\ll 100$ — the shortcut is invalid (α would be 42%, not "small").

(b) Exact quadratic: $4c{K}_a=4(10^{-4})(1.8\times 10^{-5})=7.2\times 10^{-9},K_a^2=3.24\times 10^{-10}$ $\sqrt{K_a^2+4c{K}_a}=\sqrt{7.524\times 10^{-9}}=8.674\times 10^{-5}$ $\alpha =\frac{-1.8\times 10^{-5}+8.674\times 10^{-5}}{2(10^{-4})}=\frac{6.874\times 10^{-5}}{2.0\times 10^{-4}}=0.3437=34.4\%$ $[{\text{H}}^{+}{]}_{\text{acid}}=c\alpha =1.0\times 10^{-4}\times 0.3437=3.44\times 10^{-5}\text{M}$

(c) Water check: using the $K_w$ recap above, water alone supplies $[{\text{H}}^{+}]=10^{-7}$ M. Here the acid supplies $3.44\times 10^{-5}$ M — about 340× more. So water's contribution is still negligible at this concentration. It only becomes a real correction once acid-derived $[{\text{H}}^{+}]$ drops toward $\sim 10^{-6}$ M (acid $\lesssim 10^{-6}$–$10^{-7}$ M). Lesson: the naïve formula fails first (at $c/K_a<100$), water fails much later.

Idea: in the small-α regime $\alpha =\sqrt{K_a/c}$, so equal α means equal $K_a/c$. Set them equal: $\frac{K_{a,Y}}{c_Y}=\frac{K_{a,X}}{c_X}\Rightarrow c_Y=c_X\frac{K_{a,Y}}{K_{a,X}}=0.10\times \frac{7.2\times 10^{-5}}{1.8\times 10^{-5}}=0.10\times 4=0.40\text{M}$

Verify Y at 0.40 M: ${\alpha }_Y=\sqrt{\frac{7.2\times 10^{-5}}{0.40}}=\sqrt{1.8\times 10^{-4}}=1.342\times 10^{-2}=1.34\%✓$

Real-solution caveat (ionic strength). The clean formulas above use concentrations, but the true equilibrium uses activities $a=\gamma [\cdot ]$, where $\gamma$ is an activity coefficient that drops below 1 as ions crowd. At $0.40$ M the ionic strength is high enough that $\gamma$ for the ions can fall to $\sim 0.7$–$0.8$, which effectively raises the true dissociation by a few percent versus the ideal prediction. So the "engineered α = 1.34%" is the ideal-solution value; a real 0.40 M solution would read slightly higher. Punchline: a stronger acid (bigger $K_a$) can share the same ideal α as a weaker one if you concentrate it enough — but at high $c$ remember the numbers are approximate because activity ≠ concentration.