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's the solution for question f): Step 1: Calculate the acid dissociation constant (Ka) from the given pKa. The relationship between pKa and Ka is: pKa = -_10(Ka) Given pKa = 4.7, we can find Ka: Ka = 10^-pKa Ka = 10^-4.7 Ka ≈ 1.995 × 10^-5 Step 2: Write the dissociation equilibrium for ethanoic acid and set up an ICE table. Ethanoic acid (CH_3COOH) is a weak acid, so it partially dissociates in water: CH_3COOH (aq) H^+ (aq) + CH_3COO^- (aq) Initial concentrations: [CH_3COOH] = 0.01 mol dm^-3 [H^+] = 0 [CH_3COO^-] = 0 Change in concentrations: Let x be the concentration of CH_3COOH that dissociates. [CH_3COOH] changes by -x [H^+] changes by +x [CH_3COO^-] changes by +x Equilibrium concentrations: [CH_3COOH] = 0.01 - x [H^+] = x [CH_3COO^-] = x Step 3: Use the Ka expression to solve for x, which represents [H^+]. The Ka expression is: Ka = [H^+][CH_3COO^-][CH_3COOH] Substitute the equilibrium concentrations: 1.995 × 10^-5 = (x · x)/(0.01 - x) 1.995 × 10^-5 = (x^2)/(0.01 - x) Since Ka is small and the initial concentration is relatively large, we can assume that x 0.01. This simplifies the denominator to 0.01. 1.995 × 10^-5 ≈ (x^2)/(0.01) x^2 = 1.995 × 10^-5 × 0.01 x^2 = 1.995 × 10^-7 x = sqrt(1.995 × 10^-7) x ≈ 4.466 × 10^-4 mol dm^-3 So, [H^+] ≈ 4.466 × 10^-4 mol dm^-3. Step 4: Calculate the pH. The pH is given by the formula: pH = -_10[H^+] pH = -_10(4.466 × 10^-4) pH ≈ 3.35 The pH of the ethanoic acid solution is 3.35. That's 2 down. 3 left today — send the next one.