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Edit View History Bookmarks People Window Help :-owLv21 Online teaching and IX cengagenow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator-assignment-take&takeAssignments; Reforenc Consider the titration of 100.0 mL of 0.300 M CH3NH2 by 0.150 M HCL (Kb for CH3NH2-4.4x10 ) Part 1 1 pt Calculate the pH after 0.0 mL of HCI added. pH- Part 2 Calculate the pH after 25.0 mL of HCI added. pH = Part 3 Calculate the pH after 60.0 mL of HCI added PH- Part 4 Calculate the pH at the equivalence point. 1 pt 1 pt 1 pt 1 pt 1 pt pt 1 pt 1 pt Part 5 Calculate the pH after 300.0 mL of HCI added. H- Part 6 At what volume of HCI added, does the plI-10.64? 1 pt mL Submit Show Hints ment

Explanation / Answer

part 1, ainitial

pkb = -log(4.4*10^-4) = 3.356

pH = 14-1/2(pkb-logC)

   = 14-1/2(3.356-log0.3)

   = 12.06

part2

no of mol of CH3NH2 = 100*0.3 = 30 mmol

no of mol of HCl = 25*0.15 = 3.75 mmol

pH = 14 - (pkb+log(acid/acid-base))

   = 14 - (3.356+log(3.75/(30-3.75))

   = 11.5

part 3

no of mol of CH3NH2 = 100*0.3 = 30 mmol

no of mol of HCl = 60*0.15 = 9 mmol

pH = 14 - (pkb+log(acid/acid-base))

   = 14 - (3.356+log(9/(30-9))

   = 11.0

part 4 , at equivelnce point

no of mol of CH3NH2 = 100*0.3 = 30 mmol

no of mol of HCl = 30 mmol

volume of HCl = 30/0.15 = 200 ml

concentration of salt = 30/300 = 0.1 M

pH = 7-1/2(pkb+logC)

   = 7-1/2(3.356+log0.1)

= 5.822

part 5

excess HCl concentration = (300*0.15- 30)/(400) = 0.0375 M

pH = -log(0.0375) = 1.426

part 6

no of mol of CH3NH2 = 100*0.3 = 30 mmol

no of mol of HCl = x mmol

pH = 14 - (pkb+log(acid/acid-base))

10.64 = 14 - (3.356+log(x/(30-x))

volumeof HCl = 15.07/0.15 = 100.5 ml

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