Question

You are titrating 50.0 mL of a 2.64 M benzoic acid solution with a 0.445 M potassium hydroxide solution. What is the pH of the benzoic acid solution before you begin titrating? What is the pH after 20.0 mL of the potassium hydroxide solution has been added? What volume of potassium hydroxide is needed to reach the equivalence point? What is the pH at the equivalence point? What is the pH after 332.0 mL of potassium hydroxide has been added?

Answer 1

Benzoic Acid is a week acid and its dissociation contant is k_a = 6.35 times 10^-5 O^H of Benzoic acid (2.64 M) [H^+] = squareroot k_a.c = squareroot 6.25 times 10^-5 times 2.64 = 1.28 times 106-2 M P6H = -log_10(H^+) = -log_10(1.28 times 10^-2) = -log^1.28_10 - log_10^10^-2 = -0.107 + 2 log^10_10 P^H = 1.89 \r\n b. when 20 ml of KOH (0.445M) is added, a buffer solution of (ph - CooH + ph -Cook) formed ph - cooH rightleftharpoon ph cook conc of ph-CooH After adding KOH = Conc of ph - CooH - Cons of KOH M_ = M_1 times V_1 - M_2V_2/V_1 + V_2 = 2.64 times 50 - 20 times 0.445/50 + 20 = 132 - 8.9/70 Conc of ph -cooH (remaining) = 1.76 m conc [H^+] = squareroot k_a times Conc of ph - cooH = squareroot 6.25 times 10^-5 times 1.76 = 6.8 times 10^-3 M P^H = -log_1o^[H^+] = -log 6.8 times 10^-3 = -log _10^6.8 - log^10^-3_10 = -log ^6.8_10 + 3 log^10_10 = 0.83 + 3 P^H = 2.16 \r\ n To reach equivalent point