Question

Calculate the pH when 35mL of 0.2M ammonia (Kb = 1.8 x 10^-5) is titrated with 20mL of 0.2M Hydrochloric Acid.

Answer 1

Ans

The neutralization reaction

HCl(aq]+NH3(aq]→NH+4(aq]+Cl−(aq]

hydrochloric acid, strong acid, and ammonia, a weak base,

hydrochloric acid and ammonia react in a 1:1 mole ratio.

c=nV⇒n=c⋅V

nHCl=0.2 M x 20 x 10^−3L=0.004 moles HCl

nNH3=0.2 M⋅35⋅10−3L=0.0070 moles NH3

the acid will act as a limiting reagent,

Now, since you have a 1:1 mole ratio between both reactants

So, if 0.004 moles of hydrochloric acid will react with 0.004 moles of ammonia, the reaction will produce

nNH+4=0.004 moles

of ammonium ions.

This means that after the reaction is complete, the solution will contain

nHCl=0→ completely consumed

nNH3=0.007 moles−0.004 moles=0.0030 moles NH3

nNH+4=0.0025 moles NH+4

The total volume of the solution will be

Vtotal=VHCl+VNH3

Vtotal=20 mL+35 mL=55 mL

[NH3]=[NH+4]=0.004 moles 55 x 10−3L=0.055 M

Kb=1.8⋅10−5

by the Henderson - Hasselbalch equation

pOH=pKb+log([conjugate acid][weak base])

Even without doing any calculation, you can look at this equation and say that when you have equal concentrations of weak base and conjugate acid, the pOH of the solution will be equal to pKb, since

[NH+4]=[NH3]⇒log([NH+4][NH3])=0

Since

pkb=−log(Kb)

you will have

pOH=−log(1.8⋅10−5)=4.74

The pH of the buffer will thus be

pH+pOH=14

pH=14−4.74=9.26