Calculate the pH after addition of 50.0mL of the titrant.?
B)Consider the titration of 50.0mL of 0.20M NH3 (Kb= 1.8x10^-5) with 0.20 M HNO3 . Calculate the pH after addition of 50.0mL of the titrant.
Answer:
A) Moles HBr = 80.0 x 0.25 /1000 = 0.02
Moles KOH = 40.0 x 0.50 / 1000 = 0.02
HBr and KOH are strong electrolytes , the numer of moles in the same and the solution will be neutra >> pH = 7
b) Moles NH3 = 50.0 x 0.20 /1000 = 0.01
Moles HNO3 = 50.0 x 0.20 /1000 = 0.01
NH3 + H+ >> NH4+
Total volume = 100 mL = 0.100 L
we get only NH4+ because no reactant is in excess
[NH4+] = 0.01 / 0.1 = 0.1 M
NH4+ + H2O <---> NH3 + H3O+
initial concentration
0.1
at equilibrium
0.1-x . . . .. . . . . . . . ..x . . . . . .x
The constant of this equilibrium is Kw / Kb = 5.55 x 10^-10
5.55 x 10^-10 = x^2 / 0.1 - x
x = [OH-] 7.54 x 10^-6 M
pOH = 5.13
pH = 14 - 5.13 = 8.87
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