Question

calculating the ph of a weak base titrated with a strong acid

An analytical Chemist is titrating 59.9 ml. of a 0.08300 M solution of diethylamine (CH), NH with a 0.08300 M solution of HNO,. The pl, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 68.0 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

Answer 1

no of moles of (C2H5)2NH = molarity * volume in L

                                      = 0.083*0.0599  

                                      = 0.0049717moles

no of moles of HNO3 = molarity * volume in L

                                = 0.083*0.068

                                = 0.005644moles

(C2H5)2NH(aq) + HNO3(aq) -------------> (C2H5)2NH2NO3

no of moles of excess HNO3 = 0.005644-0.0049717   = 0.0006723moles

total volume = 59.9 + 68 = 127.9ml

molarity of HNO3 = no of moles/volume in L

                          = 0.0006723/0.1279   = 0.005256M

HNO3(aq) ------------> H^+(aq) + NO3^- (aq)

0.005256M -------- 0.005256M

[H^+]   = [HNO3]   = 0.005256M

PH   = -log[H^+]

         = -log0.005256

        = 2.28 >>>>answer