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

This is the worked-examples deep dive for the parent topic on strong/weak acids and α. Before we compute anything, one promise: every symbol here is spelled out in words the first time it appears, so you can follow from line one even if you have never seen a chemistry formula.

The single relationship we lean on the whole page (built in the parent note) is:

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are tagged with the cell they cover, and together they hit every row.

Rows A–H are all covered below.

[!example] Example 1 — Cell A: ordinary weak acid, shortcut valid

Statement. Find $\alpha$ for $0.1\text{M}$ acetic acid, $K_a=1.8\times 10^{-5}$.

Forecast: Guess first — will $\alpha$ be closer to $1$ (most molecules split) or closer to $0.01$ (almost none split)? Acetic acid is the sour bit in vinegar, which you can drink, so bet on tiny.

1. Test the shortcut. $\frac{c}{K_a}=\frac{0.1}{1.8\times 10^{-5}}\approx 5556.$ Why this step? The shortcut throws away the $1-\alpha$ in the denominator. That is only safe when $\alpha$ is small. The ratio $c/K_a$ being far above $100$ is our green light.

2. Apply the shortcut. $\alpha \approx \sqrt{\frac{1.8\times 10^{-5}}{0.1}}=\sqrt{1.8\times 10^{-4}}=\mathrm{0.0134.}$ Why this step? This is the formula the ratio just told us we're allowed to use.

3. Read it as a percentage. $\alpha =1.34\%$. So about $13$ molecules in every $1000$ actually break apart.

Verify: Put $\alpha =0.0134$, $c=0.1$ back into the exact master equation: $\frac{c{\alpha }^2}{1-\alpha }=\frac{0.1\times (0.0134{)}^2}{1-0.0134}=1.82\times 10^{-5}\approx K_a.✓$ The tiny mismatch ($1.82$ vs $1.80$) is exactly the error from dropping $\alpha$ — small, as promised.

[!example] Example 2 — Cell B: what dilution does to α

Statement. Same acetic acid ($K_a=1.8\times 10^{-5}$). Compute $\alpha$ at $0.01\text{M}$ and at $1.0\text{M}$, and compare to the $0.1\text{M}$ answer above.

Forecast: As you dilute (lower $c$), does $\alpha$ go up or down? Trick: dividing by a smaller $c$ inside the square root makes the whole thing bigger.

1. Dilute case, $0.01\text{M}$. $\alpha =\sqrt{\frac{1.8\times 10^{-5}}{0.01}}=\sqrt{1.8\times 10^{-3}}=0.0424=4.24\%.$

2. Concentrated case, $1.0\text{M}$. $\alpha =\sqrt{\frac{1.8\times 10^{-5}}{1.0}}=\sqrt{1.8\times 10^{-5}}=0.00424=0.424\%.$ Why these steps? Only $c$ changed; $K_a$ is a property of the acid and never moves.

3. Look at the picture.

Dilute more → α climbs. This is Ostwald's dilution law in action, and it matches Le Chatelier thinking: spreading molecules apart makes it harder for a stray ${\text{H}}^{+}$ to find an ${\text{A}}^{-}$ and recombine, so the split state wins.

Verify: The fraction α rose from $0.42\%$ to $4.24\%$ (×10 up as $c$ went ×100 down — a ​ relationship, so $\sqrt{100}=10$ ✓). But absolute $[{\text{H}}^{+}]=c\alpha$: at $1.0\text{M}$ it is $1.0\times 0.00424=4.24\times 10^{-3}$; at $0.01\text{M}$ it is $0.01\times 0.0424=4.24\times 10^{-4}$ — so the acid still gets more acidic when concentrated even though the fraction fell. Both trends coexist. ✓

[!example] Example 3 — Cell C: the shortcut BREAKS, use the quadratic

Statement. A weak acid with $K_a=1.0\times 10^{-3}$ at concentration $c=1.0\times 10^{-3}\text{M}$. Find $\alpha$.

Forecast: Here $c/K_a=1$ — nowhere near $\gg 100$. Guess: is $\alpha$ small enough to trust the shortcut, or dangerously large?

1. Try the shortcut anyway (to see it fail). $\alpha \approx \sqrt{\frac{10^{-3}}{10^{-3}}}=\sqrt{1}=1.0(100\%).$ Why this step? To expose the danger: the shortcut spits out $\alpha =1$, which would mean complete dissociation — but then $1-\alpha =0$ and the master equation blows up. Nonsense. The shortcut is only trustworthy for tiny $\alpha$.

2. Switch to the exact quadratic. With $K_a=c=10^{-3}$: $\alpha =\frac{-K_a+\sqrt{K_a^2+4c{K}_a}}{2c}=\frac{-10^{-3}+\sqrt{(10^{-3}{)}^2+4(10^{-3})(10^{-3})}}{2\times 10^{-3}}.$ Why this step? The quadratic came from the master equation with no approximation, so it is always safe.

3. Crunch the numbers. $=\frac{-10^{-3}+\sqrt{10^{-6}+4\times 10^{-6}}}{2\times 10^{-3}}=\frac{-10^{-3}+\sqrt{5\times 10^{-6}}}{2\times 10^{-3}}=\frac{-10^{-3}+2.236\times 10^{-3}}{2\times 10^{-3}}=\mathrm{0.618.}$ So $\alpha \approx 61.8\%$ — the shortcut's "$100\%$" was off by nearly $40$ percentage points.

Verify: Plug $\alpha =0.618$, $c=10^{-3}$ into the exact master equation: $\frac{c{\alpha }^2}{1-\alpha }=\frac{10^{-3}(0.618{)}^2}{1-0.618}=\frac{10^{-3}\times 0.382}{0.382}=10^{-3}=K_a.✓$ Perfect — the ${\alpha }^2$ on top and the $(1-\alpha )$ on the bottom happen to cancel here, a neat sanity signature.

[!example] Example 4 — Cell D: the strong-acid limit

Statement. $0.1\text{M}$ HCl (a strong acid). Find $\alpha$ and pH.

Forecast: Strong acid = terrible employer, all ${\text{H}}^{+}$ quit. So α should pin at its ceiling.

1. Recognise there is no equilibrium. HCl dissociates essentially completely, so $\alpha =1$. There is no $K_a$ balancing act — the reaction arrow points one way. Why this step? Using a $K_a$ formula on a strong acid is a category error; the machinery of α, master equation, quadratic — none of it applies. This is a definitional case.

2. Compute $[{\text{H}}^{+}]$. With $\alpha =1$: $[{\text{H}}^{+}]=c\alpha =0.1\times 1=0.1\text{M}$.

3. Compute pH using $\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$ (see pH and pOH calculations): $\text{pH}=-{\log }_{10}(0.1)=\mathrm{1.0.}$

Verify: Contrast with acetic acid at the same $0.1\text{M}$: there $[{\text{H}}^{+}]=1.34\times 10^{-3}$, giving $\text{pH}=-\log (1.34\times 10^{-3})=2.87$. Ratio of $[{\text{H}}^{+}]$: $0.1/1.34\times 10^{-3}\approx 75$. Same concentration, strong acid delivers ~$75\times$ more free ${\text{H}}^{+}$. ✓

[!example] Example 5 — Cell E: a weak BASE (the mirror image)

Statement. $0.1\text{M}$ ammonia, ${\text{NH}}_3$, with base constant $K_b=1.8\times 10^{-5}$. Find $\alpha$, $[{\text{OH}}^{-}]$, and pH.

Forecast: By symmetry with acetic acid (same-size constant, same concentration), guess $\alpha$ will be almost identical to Example 1's.

1. Same formula, swap the letters. A base grabs a proton and releases ${\text{OH}}^{-}$: ${\text{NH}}_3+{\text{H}}_2\text{O}\rightleftharpoons {\text{NH}}_4^{+}+{\text{OH}}^{-}$. The ICE algebra is identical with $K_b$ in place of $K_a$: $\alpha \approx \sqrt{\frac{K_b}{c}}=\sqrt{\frac{1.8\times 10^{-5}}{0.1}}=0.0134=1.34\%.$ Why this step? Base equilibria are the mirror of acid equilibria — same math, ions on the ${\text{OH}}^{-}$ side instead of the ${\text{H}}^{+}$ side.

2. Get $[{\text{OH}}^{-}]$. $[{\text{OH}}^{-}]=c\alpha =0.1\times 0.0134=1.34\times 10^{-3}\text{M}$.

3. Convert to pH via water. First $\text{pOH}=-\log (1.34\times 10^{-3})=2.87$. Then, since $\text{pH}+\text{pOH}=14$ at $25^{\circ }\text{C}$: $\text{pH}=14-2.87=\mathrm{11.13.}$ Why this step? $[{\text{H}}^{+}]$ and $[{\text{OH}}^{-}]$ are locked together by water's self-ionisation ($K_w=10^{-14}$); knowing one gives the other.

Verify: $\alpha$ matches acetic acid exactly (same $K$, same $c$) as forecast ✓. And $\text{pH}=11.13>7$ confirms a basic solution — sanity holds.

[!example] Example 6 — Cell F: infinite dilution (degenerate limit)

Statement. For the acetic acid of Example 1, what happens to $\alpha$ as $c\to 0$ (endless dilution)? Does it reach $\alpha =1$?

Forecast: From Example 2, diluting raised α. Push dilution to the extreme — does α hit $100\%$ or stall?

1. Look at the shortcut's trend. $\alpha \approx \sqrt{K_a/c}$ grows without bound as $c\to 0$ — but α cannot exceed $1$, so the shortcut must be lying near the limit (exactly the Cell-C failure again).

2. Use the exact quadratic's limit. In $\alpha =\frac{-K_a+\sqrt{K_a^2+4c{K}_a}}{2c}$, as $c\to 0$ the numerator $\to -K_a+\sqrt{K_a^2}=-K_a+K_a=0$ and the denominator $\to 0$ too — a $0/0$. Factor carefully: for tiny $c$, $\sqrt{K_a^2+4c{K}_a}\approx K_a\sqrt{1+\frac{4c}{K_a}}\approx K_a\left(1+\frac{2c}{K_a}\right)=K_a+2c.$ So the numerator $\to -K_a+(K_a+2c)=2c$, and $\alpha \to \frac{2c}{2c}=1$. Why this step? The $0/0$ needs a careful expansion; the clean answer is $\alpha \to 1$.

3. Conclusion. At infinite dilution a weak acid dissociates completely, $\alpha \to 1$. It behaves like a strong acid — but only in the theoretical limit.

Verify: Numerically probe with $c=10^{-7}\text{M}$, $K_a=1.8\times 10^{-5}$: $\alpha =\frac{-1.8\times 10^{-5}+\sqrt{(1.8\times 10^{-5}{)}^2+4(10^{-7})(1.8\times 10^{-5})}}{2\times 10^{-7}}\approx \mathrm{0.989.}$ Already $98.9\%$ — heading to $1$ as predicted ✓. (In reality water's own ${\text{H}}^{+}$ takes over first, but the trend is real.)

[!example] Example 7 — Cell G: real-world word problem (vinegar)

Statement. Household vinegar is about $5\%$ acetic acid by mass, which works out to roughly $0.83\text{M}$. Using $K_a=1.8\times 10^{-5}$, estimate the pH and explain in one sentence why vinegar is safe to eat but concentrated HCl is not.

Forecast: Guess the pH — around $1$ (like HCl), or a milder $2$–$3$?

1. Find α with the shortcut (check first: $c/K_a=0.83/1.8\times 10^{-5}\approx 46000\gg 100$, so shortcut OK): $\alpha =\sqrt{\frac{1.8\times 10^{-5}}{0.83}}=\sqrt{2.169\times 10^{-5}}=4.657\times 10^{-3}.$
2. Get $[{\text{H}}^{+}]$. $[{\text{H}}^{+}]=c\alpha =0.83\times 4.657\times 10^{-3}=3.87\times 10^{-3}\text{M}$. Why this step? Only the dissociated fraction gives free acid; the rest sits as intact molecules doing nothing corrosive.
3. pH. $\text{pH}=-\log (3.87\times 10^{-3})=2.41$.

Verify: A $0.83\text{M}$ strong acid would give $[{\text{H}}^{+}]=0.83$, pH $\approx 0.08$ — over $200\times$ more free ${\text{H}}^{+}$. Answer sentence: vinegar is safe because only ~$0.47\%$ of its acid is ever ionised at once, so free ${\text{H}}^{+}$ stays ~$200\times$ below what the same amount of HCl would unleash. ✓

[!example] Example 8 — Cell H: exam twist, back-solve for $K_a$

Statement. A student measures that $0.05\text{M}$ of a mystery weak acid is $2.0\%$ dissociated. Find its $K_a$ (no approximation — show it exactly).

Forecast: Given $\alpha$ directly, this reverses the usual flow. Will $K_a$ come out near acetic acid's $10^{-5}$ or bigger?

1. List knowns. $c=0.05\text{M}$, $\alpha =0.02$.
2. Plug straight into the exact master equation — no shortcut needed since we know α exactly: $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.05\times (0.02{)}^2}{1-0.02}.$ Why this step? When α is given, the master equation is a direct plug-in; the quadratic is only for when α is unknown.
3. Compute. $K_a=\frac{0.05\times 4\times 10^{-4}}{0.98}=\frac{2\times 10^{-5}}{0.98}=2.04\times 10^{-5}.$

Verify: Feed $K_a=2.04\times 10^{-5}$, $c=0.05$ back through the forward quadratic and confirm α returns to $0.02$: $\alpha =\frac{-2.04\times 10^{-5}+\sqrt{(2.04\times 10^{-5}{)}^2+4(0.05)(2.04\times 10^{-5})}}{2(0.05)}=\mathrm{0.0200.}✓$ Round-trip closes.

[!recall]- Quick self-test