Cengage * Y CVLv2 I Online teaching 8 – nx CO west.cengagenowcom/irn/takeAssignm
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Cengage * Y CVLv2 I Online teaching 8 – nx CO west.cengagenowcom/irn/takeAssignmenttakeCovalentActivitydo?locator-assignment-take&takeAssignmentSessionLocator-assignment-take; Q fo: Chapter 14: EOC References Question 36 1 pt CHEMWORK Consider the titration of 100.0 ml of 0.100 M HONH, by 0.050 MHC1. (K, for HONHI, -1.1x108) Question 37 1 pt Queston 3B 1 pt Part 1 Calculate the pH after 0.0 ml, of HCl added. Question 39 1 pt Question 40 1 pt pH - 9.52 Ouestion 41 1 pt Part 2 Calculate the pH after 40.0 mL of HC1 added. Question 42 0 1 pt pH Question 43 1 pt Question 44 Part 3 Calculate the pH after 60.0 ml of HCl added. 1 pt Question 45 1 pt ph Question 46 0 1 pt Part 4 Calculate the pH at the equivalence point . Question 47 0 1 pt pH Question 48 1 pt Question 49 1 pt Part 5 Calculate the pH atter 300.0 mL of HC added. Question 60 1 pt ph Question 6 1 pt Part 6 At what volume of HCl added, does the pH = 6.04? Progress 22/61 items Due Nov 1 at 11:55 PM Previous Next Hinish Assignment Prmail Instructor Save and Fxit 8:21 PM Type here to search AE 11/1/2011Explanation / Answer
part 1, initial
pkb = -log(1.1*10^-8) = 7.96
pH = 14-1/2(pkb-logC)
= 14-1/2(7.96-log0.1)
= 9.52
part2
no of mol of HONH2 = 100*0.1 = 10 mmol
no of mol of HCl = 40*0.05 = 2 mmol
pH = 14 - (pkb+log(acid/acid-base))
= 14 - (7.96+log(2/(10-2))
= 6.642
part 3
no of mol of HONH2 = 100*0.1 = 10 mmol
no of mol of HCl = 60*0.05 = 3 mmol
pH = 14 - (pkb+log(acid/acid-base))
= 14 - (7.96+log(3/(10-3))
= 6.4
part 4 , at equivelnce point
no of mol of CH3NH2 = 100*0.1 = 10 mmol
no of mol of HCl = 10 mmol
volume of HCl = 10/0.05 = 200 ml
concentration of salt = 10/300 = 0.033 M
pH = 7-1/2(pkb+logC)
= 7-1/2(7.96+log0.033)
= 3.76
part 5
excess HCl concentration = (300*0.05- 10)/(400) = 0.0125 M
pH = -log(0.0125) = 1.9
part 6
no of mol of HONH2 = 100*0.1 = 10 mmol
no of mol of HCl = x mmol
pH = 14 - (pkb+log(acid/acid-base))
6.04 = 14 - (7.96+log(x/(10-x))
X = 5
volumeof HCl = 5/0.05 = 100. ml
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