When determining the pH of a weak acid solution, sometimes the 5\(\%\) rule can be applied to simplify the math. At what \(K_{\mathrm{a}}\) values will a \(1.0-M\) solution of a weak acid follow the 5\(\%\) rule?

The 5% rule is used to simplify the calculation of the equilibrium concentration of a weak acid and its conjugate base in an aqueous solution. According to the 5% rule, if the dissociation of a weak acid is less than 5% of the initial concentration, then the change in concentration due to dissociation can be neglected. Mathematically, this can be expressed as: \(x \leq 0.05 * C\) where x represents the change in concentration due to dissociation and C represents the initial concentration of the weak acid.

For a weak acid, HA, the equilibrium equation in an aqueous solution can be represented as: \(HA <=> H^+ + A^-\) The equilibrium constant, \(K_{\mathrm{a}}\), for this reaction is given by: \(K_{\mathrm{a}} = \frac{[H^+][A^-]}{[HA]}\) Let's assume that the change in concentration due to dissociation is x. The concentration of each species at equilibrium can then be represented as: \([H^+] = [A^-] = x\) \([HA] = C - x\) Substitute these values into the \(K_{\mathrm{a}}\) expression: \(K_{\mathrm{a}} = \frac{x^2}{C - x}\) Using the 5% rule, we can assume that x << C, so C - x ≈ C. Then, the simplified expression for \(K_{\mathrm{a}}\) is: \(K_{\mathrm{a}} \approx \frac{x^2}{C}\) Now, we need to find the range of \(K_{\mathrm{a}}\) values that will satisfy this approximation.

From Step 1, we know that x ≤ 0.05 * C. Since the initial concentration of the weak acid, C, is given as 1.0 M, we can now write: \(x \leq 0.05 * 1.0 = 0.05\) Now, we substitute this into the simplified \(K_{\mathrm{a}}\) expression from Step 2: \(K_{\mathrm{a}} \approx \frac{x^2}{C} \leq \frac{(0.05)^2}{1.0}\) Calculate this inequality: \(K_{\mathrm{a}} \leq 0.0025\) So, for a 1.0 M solution of a weak acid, the 5% rule can be applied when \(K_{\mathrm{a}}\) values are less than or equal to 0.0025.