Question

During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into molecules of its conjugate acid. For the hypothetical weak base, B we see the following: B (aq) + H30+ (aq) - BH+ (aq) + H20 (1) In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration. 2.00 mL of hydrochloric acid added to the weak base 2.00M hydrochloric acid in the buret as the titrant 15.0 mL of original 1.45 M CSHSN (pyridine) sample You are titrating 15.00 mL of a 1.45 M solution of CHN (Pyridine) with a strong acid. If you add 2.00 mL of a 2.00 M solution of hydrochloric acid, what is the final pH of the remaining weak base solution? The Kb for C5HSN (pyridine) is 1.7 X 10% final pH = M .

Answer 1

follow. CGHS NH + H₂O 19 solution: s we can write en as CGHSN + H₂otca 10.02175 mol concentrection. + change. -0.004 mol ! Equilibrium 0.01775 mol i 249) Initical - --- 140.004 mol 0.004 moli initry and of Pyridine (C6H5N) = 0.015 eit x 1.45 mol = 0.0217 5 mol con ot of HCl = 0.002 utx 2 -0.004 mol now, molcority of Coton = 0.01775 0.017 (Toted volume in liter) = 1.04211 M. morenity of Colts - 0.004 – 0.23529 m. 0.017 Now, we can use the Henderson - Hasselbach eat. poh pkb + log CCGHS not] [CH] ou 0.23529 a-log (1.7x109) L 1.04411) 2 8.769 -0.647 - 8. 122 pH = 14-POH = 121 - 8.122 = 5. 878 + loy pH = 5.878) - - -