Exercises — Ostwald's dilution law (weak acid) - α = √(Ka - C)

The simple law is trustworthy only when the computed $\alpha <0.05$. Test each:

• (a) $\alpha =\sqrt{10^7/0.1}=\sqrt{10^8}=10^4$ — a "fraction" cannot exceed 1. Nonsense → strong acid, use $[{\text{H}}^{+}]=C$. Reject.
• (b) $\alpha =\sqrt{1.8\times 10^{-5}/0.1}=\sqrt{1.8\times 10^{-4}}=0.0134$. Since $0.0134<0.05$ → valid. ✓
• (c) $\alpha =\sqrt{6.8\times 10^{-4}/10^{-4}}=\sqrt{6.8}=2.6$ — again $>1$. Nonsense → dilute weak acid ionizing so far that approximation collapses; needs the quadratic. Reject.

Only (b) is safe. The lesson: always compute $\alpha$ and then check the 5% gate before you trust it.

• $\alpha$ = fraction of acid molecules dissociated at equilibrium; a pure number, $0\le \alpha \le 1$.
• $K_a$ = acid dissociation constant, fixed at constant temperature; larger $K_a$ = stronger acid.
• $C$ = initial (analytical) concentration of acid in mol/L.

Range of $\alpha$: from $0$ (nothing ionizes) up to $1$ (complete ionization, reached only in the limit $C\to 0$).

Step 1 — naive estimate. Apply the simple law. Call this first, unrefined value ${\alpha }_0$ (the subscript "0" just labels it the zeroth, starting guess): ${\alpha }_0=\sqrt{\frac{1.8\times 10^{-4}}{0.05}}=\sqrt{3.6\times 10^{-3}}=0.0600$ Step 2 — check the gate. ${\alpha }_0=0.0600>0.05$ — it just fails the 5% rule, so setting $1-\alpha \approx 1$ is not quite honest here.

Step 3 — refine using the exact law (WHY this iteration works). The exact relation is $K_a=\frac{C{\alpha }^2}{1-\alpha }$. Solve it for $\alpha$ without throwing away $1-\alpha$: ${\alpha }^2=\frac{K_a(1-\alpha )}{C}⟹\alpha =\sqrt{\frac{K_a}{C(1-\alpha )}}.$ This is still exact, but $\alpha$ appears on both sides. The trick: put our best current guess ${\alpha }_0$ into the right side to get an improved value ${\alpha }_1$ on the left. Because ${\alpha }_0$ is already close, one pass is enough: ${\alpha }_1=\sqrt{\frac{K_a}{C(1-{\alpha }_0)}}=\sqrt{\frac{1.8\times 10^{-4}}{0.05(1-0.0600)}}=\sqrt{\frac{1.8\times 10^{-4}}{0.05\times 0.94}}=\sqrt{3.83\times 10^{-3}}=0.0619$ Use $\alpha \approx 0.0619$. $[{\text{H}}^{+}]=C\alpha =0.05\times 0.0619=3.09\times 10^{-3}\text{ M}$ $\text{pH}=-{\log }_{10}(3.09\times 10^{-3})=2.51$ Lesson: near the 5% border, replacing the discarded $1-\alpha$ with $1-{\alpha }_0$ nudges $\alpha$ up (denominator shrank below 1), so $[{\text{H}}^{+}]$ is slightly higher and pH slightly lower than the naive estimate.

"2.0% ionized" means $\alpha =0.020$. Since $\alpha <0.05$ we may use $K_a\approx C{\alpha }^2$: $K_a=C{\alpha }^2=0.20\times (0.020{)}^2=0.20\times 4.0\times 10^{-4}=8.0\times 10^{-5}$ WHY $C{\alpha }^2$ and not $C{\alpha }^2/(1-\alpha )$? Because $1-0.02=0.98\approx 1$; the correction is under 2%.

First $\text{p}{K}_a=-{\log }_{10}(1.8\times 10^{-5})=4.745$. And ${\log }_{10}C={\log }_{10}(0.10)=-1$. $\text{pH}=\frac{1}{2}(4.745-(-1))=\frac{1}{2}(5.745)=2.87$ This matches $[{\text{H}}^{+}]=\sqrt{K_{a}C}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}$, $\text{pH}=2.87$. ✓ (The shortcut is just $-{\log }_{10}$ applied to $[{\text{H}}^{+}]=\sqrt{K_{a}C}$; the square root turns into the $\frac{1}{2}$.)

Both starting $\alpha$ values stay under 0.05, so the simple law holds.

• ${\alpha }_1=\sqrt{1.8\times 10^{-5}/0.10}=0.0134$; ${\alpha }_2=\sqrt{1.8\times 10^{-5}/0.010}=0.0424$. Ratio ${\alpha }_2/{\alpha }_1=0.0424/0.0134=3.16=\sqrt{10}$. So a 10× dilution raises $\alpha$ by $\sqrt{10}$.
• $[{\text{H}}^{+}{]}_1=\sqrt{K_a{C}_1}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}$; $[{\text{H}}^{+}{]}_2=\sqrt{1.8\times 10^{-7}}=4.24\times 10^{-4}$. Ratio $=4.24\times 10^{-4}/1.34\times 10^{-3}=0.316=1/\sqrt{10}$. So $[{\text{H}}^{+}]$ fell by $\sqrt{10}$.

The paradox resolved. $\alpha \propto 1/\sqrt{C}$ rises, but $[{\text{H}}^{+}]\propto \sqrt{C}$ falls. Each molecule is more likely to ionize, yet there are far fewer molecules per litre — the count of free H⁺ drops.

The figure below makes this a single picture. The horizontal axis is concentration $C$ on a log scale (concentrated on the right, dilute on the left). The magenta curve is $\alpha$: read it against the left axis — as you move left (dilute), it climbs and eventually plateaus at 1 because a fraction cannot exceed 100%. The violet curve is $[{\text{H}}^{+}]$: read it against the right axis — it slides downward as you dilute. The two orange dots mark our 0.10 M and 0.010 M points on the magenta ($\alpha$) curve; the navy squares mark the same concentrations on the violet ($[{\text{H}}^{+}]$) curve. One curve up, the other down, at the very same concentrations — that is the paradox, drawn. This opposed behaviour is exactly Le Chatelier's Principle: dilution adds solvent, so the system shifts toward the side with more dissolved particles (2 ions H⁺ + A⁻ beat 1 molecule HA), raising the fraction ionized — yet spreading everything thinner still lowers the absolute ion count.

Since the naive $\alpha =0.825$ badly violates the 5% rule, use the exact form: $K_a=\frac{C{\alpha }^2}{1-\alpha }\Rightarrow 6.8\times 10^{-4}=\frac{0.001{\alpha }^2}{1-\alpha }$ Multiply out: $0.001{\alpha }^2=6.8\times 10^{-4}(1-\alpha )$, i.e. $0.001{\alpha }^2+6.8\times 10^{-4}\alpha -6.8\times 10^{-4}=0.$ Divide by 0.001: ${\alpha }^2+0.68\alpha -0.68=0$. Quadratic formula (take the positive root): $\alpha =\frac{-0.68+\sqrt{0.68^2+4(0.68)}}{2}=\frac{-0.68+\sqrt{3.1824}}{2}=\frac{-0.68+1.7839}{2}=0.552$ True $\alpha =0.55$, not 0.83. The simple law overstated ionization by ~50%. At such dilution HF is genuinely half-ionized, and the "$1-\alpha$" depletion of undissociated HA can no longer be ignored.

Sodium acetate is a salt that dissolves completely, so it dumps A⁻ into the solution before any acetic acid ionizes. Now A⁻ starts at 0.10 M. Let $x=[{\text{H}}^{+}]$ produced by the tiny bit of acid that still ionizes. Equilibrium concentrations: $[\text{HA}]\approx 0.10-x\approx 0.10$, and $[{\text{A}}^{-}]\approx 0.10+x\approx 0.10$ (both because $x$ will turn out tiny). Substitute into the exact equilibrium expression: $K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}=\frac{x(0.10)}{0.10}=x\Rightarrow x=[{\text{H}}^{+}]=K_a=1.8\times 10^{-5}\text{ M}$ $\text{pH}=-{\log }_{10}(1.8\times 10^{-5})=4.74$ Where the Henderson–Hasselbalch equation comes from (derived here, not assumed). Take $-{\log }_{10}$ of $K_a=[{\text{H}}^{+}]\frac{[{\text{A}}^{-}]}{[\text{HA}]}$: $-{\log }_{10}{K}_a=-{\log }_{10}[{\text{H}}^{+}]-{\log }_{10}\frac{[{\text{A}}^{-}]}{[\text{HA}]}⟹\text{p}{K}_a=\text{pH}-{\log }_{10}\frac{[{\text{A}}^{-}]}{[\text{HA}]},$ i.e. $\text{pH}=\text{p}{K}_a+{\log }_{10}\frac{[{\text{A}}^{-}]}{[\text{HA}]}$ — that is the Henderson-Hasselbalch Equation. Here $[{\text{A}}^{-}]=[\text{HA}]=0.10$, so ${\log }_{10}1=0$ and $\text{pH}=\text{p}{K}_a=4.74$, matching our direct calculation. New $\alpha =[{\text{H}}^{+}]/C=1.8\times 10^{-5}/0.10=1.8\times 10^{-4}$. Compare: pure acid had $\alpha =0.0134$. Adding the common ion crushed $\alpha$ by a factor of ~75. This is the Common Ion Effect: the pre-loaded A⁻ pushes the equilibrium HA ⇌ H⁺ + A⁻ back toward undissociated HA, suppressing ionization — and pH jumps from 2.87 to 4.74.

Only ionized molecules carry current. At infinite dilution every molecule is ionized, giving the ceiling conductivity ${\Lambda }_m^{\infty }$; at concentration $C$ only the fraction $\alpha$ is ionized, giving ${\Lambda }_m$. So the measured-to-ceiling ratio is the ionized fraction: $\alpha ={\Lambda }_m/{\Lambda }_m^{\infty }$. This is the practical bridge described in Conductivity of Solutions — it lets a bench conductivity meter read off $\alpha$ directly. Substitute into the exact law: $K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{C{\left(\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}\right)}^2}{1-\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}}=\frac{C{\Lambda }_m^2}{{\Lambda }_m^{\infty }\left({\Lambda }_m^{\infty }-{\Lambda }_m\right)}$ This is Ostwald's dilution law in its conductivity form — the version historically used to measure $K_a$. No approximation was made, so it works even for moderately ionized acids.

$\alpha =0.50$ is nowhere near small, so the simple law is forbidden — use the exact form solved for $C$: $K_a=\frac{C{\alpha }^2}{1-\alpha }\Rightarrow C=\frac{K_a(1-\alpha )}{{\alpha }^2}=\frac{1.8\times 10^{-5}(1-0.50)}{0.50^2}=\frac{1.8\times 10^{-5}\times 0.50}{0.25}$ $C=\frac{9.0\times 10^{-6}}{0.25}=3.6\times 10^{-5}\text{ M}$ So you must dilute to $3.6\times 10^{-5}$ M — extremely dilute — to reach half ionization. $[{\text{H}}^{+}]=C\alpha =3.6\times 10^{-5}\times 0.50=1.8\times 10^{-5}\text{ M},\text{pH}=-{\log }_{10}(1.8\times 10^{-5})=4.74$ Insight: at $\alpha =0.5$ we again land on $\text{pH}=\text{p}{K}_a$ — but for a completely different reason than Ex 4.1 (here it's extreme dilution, there it was equal salt/acid). This $\alpha \approx 0.5$ region is also where Buffer Capacity would be maximal if the conjugate base were present: a buffer resists pH change best when HA and A⁻ are equal, so a chemist deliberately targets this ratio when mixing a buffer.

Treat the first proton loss as its own weak-acid equilibrium (this is the standard move for Polyprotic Acids: each proton leaves in its own step with its own constant): ${\alpha }_0=\sqrt{\frac{K_{a1}}{C}}=\sqrt{\frac{1.0\times 10^{-3}}{0.10}}=\sqrt{1.0\times 10^{-2}}=0.10$ Check the gate: ${\alpha }_0=0.10>0.05$ — refine once (same iteration as Ex 2.1, using $1-{\alpha }_0=0.90$): ${\alpha }_1=\sqrt{\frac{K_{a1}}{C(1-{\alpha }_0)}}=\sqrt{\frac{1.0\times 10^{-3}}{0.10\times 0.90}}=\sqrt{1.11\times 10^{-2}}=0.105$ So ${\alpha }_1\approx 0.11$ (about 11%). Why stage-by-stage? $K_{a2}/K_{a1}=10^{-5}$ — the second ionization is a hundred-thousand times weaker, so essentially all the H⁺ comes from step 1. We apply Ostwald independently to $K_{a1}$; the second step contributes negligibly to $[{\text{H}}^{+}]$. Treating them separately is valid because the two constants are so far apart.

From the simple law, $\alpha =\sqrt{K_a/C}$, so $\alpha \propto C^{-1/2}$. Replacing $C\to C/100$: $\frac{{\alpha }_{\text{new}}}{{\alpha }_{\text{old}}}=\sqrt{\frac{C}{C/100}}=\sqrt{100}=10.$ And $[{\text{H}}^{+}]=\sqrt{K_{a}C}\propto C^{1/2}$, so $\frac{[{\text{H}}^{+}{]}_{\text{new}}}{[{\text{H}}^{+}{]}_{\text{old}}}=\sqrt{\frac{C/100}{C}}=\sqrt{\frac{1}{100}}=\frac{1}{10}.$ General rule: dilute by a factor $f$ → $\alpha$ multiplies by $\sqrt{f}$ and $[{\text{H}}^{+}]$ divides by $\sqrt{f}$ (equivalently, pH rises by $\frac{1}{2}{\log }_{10}f$). This holds only while the diluted $\alpha$ stays below 0.05.

Before trusting any Ostwald answer: Did I compute α and confirm it is below 0.05? ::: If not, use the quadratic or iterate. Is a conjugate salt present? ::: If yes, use Henderson–Hasselbalch, not ½(pKa − log C). Did I distinguish fraction (α) from count ([H⁺])? ::: Dilution raises α but lowers [H⁺].