This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions: d) i) A weak acid is an acid that only partially dissociates (ionizes) in an aqueous solution, meaning it does not release all of its hydrogen ions. ii) To calculate the pH of a 0.25 mol dm^-3 solution of ethanoic acid (CH_3COOH) with K_a = 1.8 × 10^-5 mol dm^-3: Step 1: Write the dissociation equilibrium for ethanoic acid. CH_3COOH(aq) CH_3COO^-(aq) + H^+(aq) Step 2: Set up an ICE (Initial, Change, Equilibrium) table for concentrations. Let x be the concentration of H^+ ions at equilibrium. | Species | Initial (mol dm^-3) | Change (mol dm^-3) | Equilibrium (mol dm^-3) | | :------------- | :---------------------- | :--------------------- | :-------------------------- | | CH_3COOH | 0.25 | -x | 0.25 - x | | CH_3COO^- | 0 | +x | x | | H^+ | 0 | +x | x | Step 3: Write the expression for the acid dissociation constant, K_a. K_a = [CH_3COO^-][H^+][CH_3COOH] Step 4: Substitute the equilibrium concentrations into the K_a expression. 1.8 × 10^-5 = ((x)(x))/(0.25 - x) Step 5: Since K_a is small, assume x 0.25, so 0.25 - x ≈ 0.25. 1.8 × 10^-5 = (x^2)/(0.25) Step 6: Solve for x. x^2 = 1.8 × 10^-5 × 0.25 x^2 = 4.5 × 10^-6 x = sqrt(4.5 × 10^-6) x = 0.002121 mol dm^-3 This value of x represents [H^+]. Step 7: Calculate the pH. pH = -[H^+] pH = -(0.002121) pH ≈ 2.67 The pH of the solution is 2.67. e) i) A buffer solution is a solution that resists significant changes in pH when small amounts of an acid or a base are added to it. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. ii) When a small amount of HCl (a strong acid) is added to the buffer solution (a mixture of CH_3COOH and CH_3COONa), the added H^+ ions from HCl react with the conjugate base, ethanoate ions (CH_3COO^-), to form the weak acid, ethanoic acid (CH_3COOH). This consumes the added H^+ ions, preventing a large drop in pH. CH_3COO^-(aq) + H^+(aq) → CH_3COOH(aq) 3 done, 2 left today. You're making progress.