Consider the titration of of 60.0 mL of 0.500 M NH4OH (Kb= 2.0x10^-5) with 0.500 M HNO3. Determine the pH at the following volume additions of the acid: (a) 60.0 mL and (b) 65.0 mL.
a) 60.0ml
HNO3 + NH3 ------> NH4+ + NO3-
1:1molar reaction
given moles of NH3 = ( 0.500mol/1000ml) × 60.0ml = 0.03mol
moles of HNO3 required to react with 0.03moles of NH3 = 0.03mol
Volume of HNO3 required = (1000ml /0.500mol) × 0.03mol = 60ml
Therefore ,
equilvalence point = 60.0ml
at equivalence point , [NH4+] = 0.500M / 2= 0.250M
NH4+ partly dissociate
NH4+ (aq) <-------> NH3(aq) + H+ (aq)
Ka = [NH3][H+]/[NH4+]
Ka =Kw/Kb = 1.00×10-14/2.0 ×10-5 = 5.0 ×10-10
at equilibrium
[NH4+] = 0.250 - x
[NH3] = x
[H+] = x
so,
x2/ ( 0.250 - x) = 5.0 ×10-10
solving for x
x = 0.00001118
[H+]= 0.00001118M
pH = -log[H+]
pH = -log(0.00001118)
pH = 4.95
b)
Excess volume of HNO3 added = 5.0ml
Excess moles of HNO3 added = ( 0.500mol/1000ml) × 5.0ml = 0.0025mol
Total volume = 125ml
[HNO3] = (0.0025mol/125ml)×1000ml = 0.0200M
[H+] = 0.0200M
pH = -log(0.0200M)
pH = 1.70
Consider the titration of of 60.0 mL of 0.500 M NH4OH (Kb= 2.0x10^-5) with 0.500 M...
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