Question

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.

Answer 1

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