Question

Calculate the change in pH when 4.00 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). (Could you please show work, thank you!

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 4/1000 = 4 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 ( 4 X 10^-4) than NH3 ( 0.01) so HCl is the limiting reagent

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

new no.of moles of NH4Cl = 0.01 + 4 X 10^-4 = 0.0104
new no.of moles of NH3 = 0.01 - 4 X 10^-4 = 0.0096

total volume = 4 + 100 = 104 ml = 0.104 L

new [NH3] = 0.0096/0.104 = 0.0923 M
[NH4Cl] = 0.0104/0.104 = 0.1 M

new pOH = 4.745 + log 0.1/0.0923

pOH = 4.745 + log 1.083

pOH = 4.745 + 0.0346 = 4.7796

new pH = 14 - 4.7796 = 9.2204

change in pH = 9.255 - 9.2204 = 0.0346