CH 17 #42 – weak base titration

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Presentation transcript:

CH 17 #42 – weak base titration CH 17 #42 a– weak base ionization in pure water. H20 + NH3  NH4+ + OH- THIS IS BEFORE THE TITRATION, NO H+ ADDED YET. Kb = [NH4+][OH-] [NH3] 1.8 * 10-5 = [X][X] [0.030-X] 1.8 * 10-5 = [X][X] [0.030] X = 7.348 * 10-4 = [OH-] ; pOH- =3.14 ; pH = 10.86

NEUTRAIZATION STOICHIOMETRY CH 17 #42 b– weak base TITRATION AFTER 0.010 L OF H+ IS ADDED TO THE BASE SOLUTION IN PART b. H+ ADDED = 0.0250 M * 0.0100L = 2.5 * 10-4 mol NH3 INITIAL = 0.030 M ( 9.0 * 10-4 MOLE) Kb = [NH4+][OH-] [NH3] NEUTRAIZATION STOICHIOMETRY (MOLES) NH3 H+  NH4+ 9.0*10-4 2.5*10-4 -X X 6.5*10 -4 2.5*10-4 MOL /0.040L=6.25*10-3 MOLAR 1.8 * 10-5 = [6.25*10-3][X] [0.01625] X = 4.6*10-5 MOLAR : pH = 9.66 6.5*10-4/0.040L = 0.01625

NEUTRAIZATION STOICHIOMETRY CH 17 #42 c– weak base TITRATION AFTER 0.020 L OF H+ IS ADDED TO THE BASE SOLUTION IN PART c. H+ ADDED = 0.0250 M * 0.0200L = 5.0 * 10-4 mol NH3 INITIAL = 0.030 M ( 9.0 * 10-4 MOLE) Kb = [NH4+][OH-] [NH3] NEUTRAIZATION STOICHIOMETRY (MOLES) NH3 H+  NH4+ 9.0*10-4 5.0*10-4 -X X 4.0*10 -4 5.0*10-4 MOL /0.050L=0.010 MOLAR 1.8 * 10-5 = [0.010][X] [0.008] OH- = X = 1.44*10-5 MOLAR : pH = 9.19 4.0*10-4/0.050L = 0.008M

NEUTRAIZATION STOICHIOMETRY CH 17 #42 d– weak base TITRATION AFTER 0.0350 L OF H+ IS ADDED TO THE BASE SOLUTION IN PART d. H+ ADDED = 0.0250 M * 0.0350L = 8.75 * 10-4 mol NH3 INITIAL = 0.030 M ( 9.0 * 10-4 MOLE) Kb = [NH4+][OH-] [NH3] NEUTRAIZATION STOICHIOMETRY (MOLES) NH3 H+  NH4+ 9.0*10-4 8.75*10-4 -X X 0.25*10 -4 8.75*10-4 MOL /0.065L=0.01346 MOLAR 1.8 * 10-5 = [0.01346][X] [4.0*10-4] OH- = X = 5.349*10-7 MOLAR : pH = 7.73 0.25*10-4/0.065L = 4.0*10-4M

NEUTRAIZATION STOICHIOMETRY WILL NOW UNDERGO HYDROLYSIS CH 17 #42 e– weak base TITRATION AFTER 0.0360 L OF H+ IS ADDED TO THE BASE SOLUTION IN PART e. H+ ADDED = 0.0250 M * 0.0360L = 9.0 * 10-4 MOLE NH3 INITIAL = 0.030 M ( 9.0 * 10-4 MOLE) ♪ NOTE: THE MOLES OF H+ ADDED = THE MOLES OF BASE INITIAL, YOU ARE AT EQUIVALENCE! NEUTRAIZATION STOICHIOMETRY (MOLES) NH3 H+  NH4+ 9.0*10-4 -X X WILL NOW UNDERGO HYDROLYSIS

CH 17 #42 e(continued)– weak base TITRATION AT EQUIVALANCE CH 17 #42 e(continued)– weak base TITRATION AT EQUIVALANCE. H2O + NH4+  NH3 + H+ NH4+   NH3 H+ 9.0*10-4 -X X 9.0*10-4 -X Ka = [NH3][H+] [NH4+] 5.56*10-10 = [X][X] [9.0*10-4mol/0.660 L] 5.56*10-10 = [X][X] [.01364 M] X = 2.75 * 10-10 M H+ pH = 5.56