Question

50.00 mL of 0.1000 M ammonia is titrated with 0.1000 M HCl.What is the pH of the solution when half the volume of HCl required for full reaction, is added? Kb of ammonia is 1.75 *10-5

Answer 1

for full reaction

Ma x Va = Mb x Vb

0.1 x Va = 0.1 x 50

Va = 50

so

for full reaction

50 ml of HCl is required

given half reaction

so

25 ml of HCl is added

now

we know that

moles = molarity x volume (L)

so

moles of NH3 = 0.1 x 50 x 10-3 = 5 x 10-3

moles of HCl added = 0.1 x 25 x 10-3 = 2.5 x 10-3

now

the reaction is

HCl + NH3 --- NH4+ + Cl-

now

moles of NH3 reacted = moles of HCL added = 2.5 x 10-3

so

moles of NH3 left = 5 x 10-3 - 2.5 x 10-3 = 2.5 x 10-3

now

moles of NH4+ formed = moles of HCl added = 2.5 x 10-3

now

HCl is totally consumed

only NH3 and NH4+ are in the solution

they form a buffer

and for buffers

pOH = pKb + log [coonjugate acid / base ]

also

pKb = -log Kb

so

pOH = -log Kb + log [NH4+ /NH3]

pOH = -log 1.75 x 10-5 + log [ 2.5 x 10-3 / 2.5 x 10-3 ]

pOH = 4.757 + 0

pOH = 4.757

now

pH = 14 - pOH

so

pH = 14 - 4.757

pH = 9.243

so

pH of the solution is 9.243