A ALEKS: Philip Potapchuk AChernist Titrates 80 0mL Chegg Study Guided x Maddy..

Question

A ALEKS: Philip Potapchuk AChernist Titrates 80 0mL Chegg Study Guided x Maddy.. × C https://www-awb.aleks.com/alekscgi/x/Isl.ex https://www-a b aleks.co aleks gw s e e/10 u IgNskr7 P3 H IB a n-l 6 P KA 3al eABbNCC im 1 M 3 47r7 2PI O ACIDS, BASES AND AQUEOUS EQUILIBRIA -Calculating the pH at equivalence of a titration A chemist titrates 190.0 mL of a 0.2122 M lidocaine (Ci,H21NONH) solution with 0.5078 M HBr solution at 25 C. Calculate the pH at equivalence. The pk, of lidocaine is 7.94. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. PH - Check

Explanation / Answer

Moles of lidocaine = molarity X volume = 0.2122 X 0.19 = 0.0403

so moles of HBr required = 0.0403 (to reach equivalence point)

Volume of HBr added = moles / molarity = 0.0403 / 0.5078 = 0.0794 Litres = 79.4 mL

Total volume = 79.4 + 190 = 269.4 mL

Moles of salt formed = 0.0403

Molarity of salt = 0.0403 / 0.269 = 0.149 molar = 0.15 molar (BNH2+)

                     BNH2+ + H2O --> 2BNH + H3O+

initial              0.15                       0           0

change             -x                         +x        +x

Equilibrium      0.15-x                     x           x

pKb = 7.94

Kb = 1.15 X 10^-8

Ka = Kw / Kb = 10^-14 / 1.15 X 10^-8 = 8.69 X 10^-7 = x^2 / (0.15-x)

8.69 X 10^-7 = x^2 / 0.15

x^2 = 1.304 X 10^-7

x = 3.61 X 10^-4= H+

pH = -log[H+] = 3.44

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