Question

a) Calculate the pH of a buffer system that contains o.200 M of NH3(aq) and o.200 M of NH4Cl(aq). The Kb of NH3(aq) is 18 x 10-5. Consult Textbook Numerical Answer pH b) Calculate the change in pH when 6.00 mL of 100 M HCl is added to 100.0 mL of the buffer solution. Numerical Answer: c) Calculate the change in pH when 6.00 mL of o.100 M NaOH is added to 100.0 mL of the buffer solution. Numerical Answer:

Answer 1

Given [NH_3] = 0.2 M = [Base] {Salt] = [NH_4Cl] = 0.2 M K_b = 1.8 times 10^-5 from Henderson - Hasselbalch equation p^DH = p^k_b + log (Salt)/[Base] equation (1) p^k_b = -log k_b = -log 1.8 times 10^-5 = 4.7447 From eq (1) p^OH = 4.7447 + log 0.2/0.2 = 4.7447 + log 1 = 4.7447 + 0 = 4.7447 Now p^H + p^OH = 14 p^H = 14 - p^OH = 14 - 4.7447 = 9.2553 6.00 ml of 0.1m Hcl added to 100.0 ml of Buffer No. of milimoles & Hcl added = 6 times 0.1 = 0.6 m moles Initially the equilibrium in buffer NH_3 + H_2O rightwardsharpoonoverleftwardsharpoon NH^+_4 + OH^- \r\n Again NH_3 + H_2O rightwardsharpoonoverleftwardsharpoon NH^+_4 + 3H^- when 0.6 m moles of acid is added, 0.6 m moles of H^+ ions are produced. These combine with 0.6 m moles of OH^- to from H_2O and the equilibrium is shifted of right (Le chatelier's principle). So,