Question

1)Consider the titration of 50.0 mL of 0.20 M  NH3 (Kb=1.8

Answer 1

PartB. Consider the titration of 50.0mL of 0.20M NH3 (Kb= 1.8*10^-5) with 0.20M of HNO3 . Calculate the pH after addition of 50.0mL of the titrant.

this reaction is @ its equivalence point, it has made
0.050 litres @ 0.2 mol/litre = 0.010 moles of NH4+ ion

0.010 moles NH4+ ion / 0.100 litres total = 0.1 molar NH4+

this NH4+ does a hydrolysis in the water:

NH4+ & H2O --> NH4OH & H+

Khydrolysis = Kwater / Kb = 1e-14 / 1.8e-5 = 5.56 e-10

5.56e-10 = [NH4OH] [ H+] / [NH4+]

5.56e-10 = [x] [ x] / [0.1 molar]

x= H+ = 7.45 e-6 molar

your answer: pH = 5.13
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Part C. A 30.0mL volume of 0.50M CH3COOH (Ka=1.8*10^-5) was titrated with 0.50M of NaOH . Calculate the pH after addition of 30.0mL of NaOH.

equal amounts of moles, this too is at its equalence point

0.50 Molar acetic acid has been jeutralized & its volume doubled,... which has cut the concentration of acetate in half to 0.25 Molar

acetate also does a hydrolysis in water:

C2H3O2- & water --> HC2H3O2 & OH-

Khydrolysis = [HC2H3O2] [OH-] / [C2H3O2-]

5.56e-10 = [x] [x] / [0.25 molar]

x = [OH-] = 1.18e-5

pOH = 4.93

your answer: pH = 9.07