Q.Using the same weak acid as in the worked example above ($K_a = 1.8 \times 10^{-5}$), calculate its degree of ionization $\alpha$ in a more dilute, $0.01\ \text{M}$ solution, and comment on how $\alpha$ changes on dilution.

[!TLDR] Apply $\alpha \approx \sqrt{K_a/C}$ at the new, ten-fold lower concentration. [!ANSWER] $\alpha \approx 0.0424$, i.e. about $4.24\%$ — roughly $3.2\times$ larger than at …

$\alpha \approx \sqrt{K_a/C} = \sqrt{(1.8\times10^{-5})/(0.01)} = \sqrt{1.8\times10^{-3}}$. Since $\sqrt{1.8} \approx 1.342$ and $\sqrt{10^{-3}} \approx 0.03162$, $\alpha \approx 1.342 \times 0.03162 \approx 0.04243$. Comparing with the earlier $0.1\ \text{M}$ result ($\alpha \approx 0.0134$), the ratio $0.0424/0.0134 \approx 3.16 \approx \sqrt{10}$ — exactly the factor expected from $\alpha \propto 1/\sqrt{C}$ when …

Recompute $\alpha \approx \sqrt{K_a/C}$ at the new concentration, and compare the ratio with $\sqrt{10}$ …

A frequent error is to expect $\alpha$ to scale by a full factor of $10$ (matching the dilution factor) rather than by $\sqrt{10}$, whi …