Question

Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

NH3 Kb=1.8x10^-5

Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

find pH initially when only NH3 and NH4Cl were present ....as it is a basic buffer mixture ...

so pOH = pKb + log [NH4Cl]/[NH3]

pKb for NH3 = 4.745

so pOH = 4.745 + log 0.1/0.1
pOH = 4.745 + log 1
pOH = 4.745 + 0 = 4.745

and pH = 14 - pOH = 14 - 4.745 = 9.255

initial no.of moles of NH3 = molarity X volume in litres = 0.1 X 100/1000 = 0.01
no.of moles of NH4Cl = 0.1 X 100/1000 = 0.01
no.of moles of HCl = 0.1 X 3/1000 = 3 X 10^-4

now when you add HCl ...following reaction will take place

NH3 + HCl -------> NH4Cl

so more NH4Cl will be formed

as HCl is present in lesser amount ( 3 X 10^-4) than NH3 ( 0.01) so HCl is the limiting reagent

no.of moles of NH4Cl formed = 3 X 10^-4

new no.of moles of NH4Cl = 0.01 + 3 X 10^-4 = 0.0103
new no.of moles of NH3 = 0.01 - 3 X 10^-4 = 0.0097

total volume = 3 + 100 = 103 ml = 0.103 L

new [NH3] = 0.0097/0.103 = 0.0941 M
[NH4Cl] = 0.0103/0.103 = 0.1 M

new pOH = 4.745 + log 0.1/0.0941

pOH = 4.745 + 0.0264 = 4.771

new pH = 14 - 4.771 = 9.229

change in pH = 9.255 - 9.229 =0.026?

2) no.of moles of NaOH = 0.1 X 3*1000 = 3 X 10^-4

this time this reaction will take place :-

NH4Cl + NaOH -------> NH3 + NaCl + H2O

so this time NH4Cl is consumed and NH3 is formed

new no.of moles of NH4Cl = 0.01 - 3 X 10^-4 = 0.0097
and no.of moles of NH3 = 0.01 + 3 X 10^-4 = 0.0103

new [NH4Cl] = 0.0097/0.103= 0.0941 M
[NH3] =0.0103/0.103 = 0.1 M

pOH = 4.745 + log 0.0941/0.1


pOH = 4.745 -0.0264 = 4.7186

so new pH = 14 - 4.7186 = 9.2814

change = 9.2814 - 9.255 = 0.0264