Question

ACIDS AND BASES Calculating the pH of a weak base titrated with a strong An analytical Chemist is titrating 140.7 mL of a 0.6100 M solution of diethylamine (CH), NH with a 0.1700 M solution of HNO. The pk, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 180.5 ml of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. pH-0 X 5 ?

Answer 1

weak base: (C2H5)2NH strong acid: HNO3 [ (C2H5)2NH ] :0.610 M [ HNO3 0.170 M pH= ? pkb (C2H5)2NH - 2.89 Calculating initial number of moles: molariy of (C2H5)2NH = 0.610 M volume of (C2H5)2NH = 140.7 ml = 0.1407 L number of moles of (C2H5)2NH = ? number of moles of (C2H5)2NH = Molarity x volume number of moles of (C2H5)2NH = 0.610 * 0.1407 number of moles of (C2H5)2NH = 0.086 mol Calculating number of moles added: molariy of HNO3 = 0.170 M volume of HNO3 = 180.5 ml = 0.1805 L number of moles of HNO3 = ? number of moles of HNO3 = Molarity x volume number of moles of HNO3 = 0.170 * 0.1805 number of moles of HNO3 = 0.031 mol

Construct ICA table to obtain their equilibrium concentrations: Reaction (C2H5)2NH + HNO3 → (C2H5)2NH2NO3 concentration ( mol ) 0.086 0.031 0.000 change ( mol ) -0.031 - 0.031 +0.031 equilibrium ( mol ) 0.055 0.000 0.031 Excess Calculating pH: pH = 14 - pKb + log [ salt ] [ base ) pH = 14 - pH = 14 - [ 2.89 + log [ 0.031 / 1 + log [ 0.055 ] pH = 11.36