Prove the statement that when the concentration of a weak acid HA decreases by a factor of \(10,\) its percent ionization increases by a factor of \(\sqrt{10}\). State any assumptions.

A weak acid is one that does not fully ionize in a solution. Let's denote our weak acid as HA. Its ionization in water can be represented as follows:\[HA \leftrightarrow H^+ + A^-\]Since this reaction does not proceed to completion, we represent the acid dissociation constant \(Ka\) as \[Ka = \frac{[H^+][A^-]}{[HA]}\]

Let's denote the initial concentration of our weak acid as \(C\) and let \(x\) be the small amount that ionizes. Then the ionization can be represented as follows where \([H^+] = [A^-] = x\) and \([HA] = C - x\).\nThe \(Ka\) can now be written as\n\[Ka = \frac{x^2}{C - x}\]\nThe percent ionization is given by \(\frac{x}{C}\) and can be ignored in the equation for \(Ka,\) since \(x\) is exceedingly small for weak acids.

Based on the assumptions made earlier, we will use the simplified equation for conception:\[Ka = \frac{x^2}{C}\]\nNow, suppose we decrease the initial concentration to \(C/10\). Let's say the new extent of ionization is \(y\): then the equation for \(Ka\) becomes \[Ka = \frac{y^2}{C/10}\]\nWe need to prove that percent ionization \(y/(C/10)\) is equal to \(\sqrt{10}\) times the old percent ionization \(x/C\). To prove this, We can divide the second equation by the first one, resulting in:\[ \frac{y^2/C}{x^2/C} = 10 \]That simplifies to:\[ \frac{y}{x} = \sqrt{10} \]Hence, proving that when the concentration of the weak concentration decreases by a factor of 10, its percent ionization increases by a factor of \(\sqrt{10}\) as required by the original problem statement.