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.
ScanSolve AI Answer
Here are the solutions for parts (a), (b), and (c) of Question 2. The titration involves a weak base, triethylamine (B), with a strong acid, HCl. The reaction is: B(aq) + H^+(aq) BH^+(aq) Given: Initial volume of weak base = 20.00 mL = 0.02000 L Initial concentration of weak base = 0.1000 M Initial moles of weak base = 0.02000 L × 0.1000 M = 0.002000 mol Concentration of titrant (HCl) = 0.1000 M K_b of base = 5.2 × 10^-4 pK_b = -(5.2 × 10^-4) ≈ 3.284 Part (a): Find the pH during the titration. i. 0.00 mL HCl added: This is a solution of a weak base. The equilibrium is: B(aq) + H_2O(l) BH^+(aq) + OH^-(aq) Let [OH^-] = x. K_b = ([BH^+][OH^-])/([B]) 5.2 × 10^-4 = (x^2)/(0.1000 - x) Using the quadratic formula (x^2 + (5.2 × 10^-4)x - (5.2 × 10^-5) = 0): x = -5.2 × 10^-4 + sqrt((5.2 × 10^-4))^2 - 4(1)(-5.2 × 10^-5)2 x = -5.2 × 10^-4 + sqrt(2.704 × 10^-7) + 2.08 × 10^-42 x = -5.2 × 10^-4 + sqrt(2.082704 × 10^-4)2 x = -5.2 × 10^-4 + 0.014432 = 0.006955 M So, [OH^-] = 0.006955 M. pOH = -(0.006955) ≈ 2.1578 pH = 14 - pOH = 14 - 2.1578 = 11.8422 The pH is $11.