Question

A 25.00 mL sample of .0056 M triethylamine (N(C2H5)3 pKb= 3.25) is titrated with standardized .0056 M HCL solution. What is the pH of the solution after 12.5 mL of HCl has been added? (Hint- half way to equivalence point)n

Answer 1

Addition of 12.5 mL of HCI: Number of moles of (CH3)2N = M*V = 0.0056 M* 25.00 mL = 0.14 mmol Number of moles of HCl = M*V = 0.0056 M* 12.50 mL = 0.7 mmol Net moles of (CH3)2N = 0.14 - 0.7 = 0.7 mmol mmol (C2H3)3 + HCI Kb = (C2H3)2NH+Cl- Initial Change 0.14 -0.7 0.7 -0.7 +0.7 Equillibrium 0.7 According to Henderson's equation 0.7 07 pOH = pK, +log = 3.25+ log =3.25+0 [base] pOH = 3.25 pH = 14 - POH = 14 – 3.25 = 10.75