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?

1) 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).
2) Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
The best explanation gets 10 points! Thanks in advance!
Posted: May 20, 2013

Best Answer

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
2) no.of moles of NaOH = 0.1 X 4/1000 = 4 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 - 4 X 10^-4 = 0.0096
and no.of moles of NH3 = 0.01 + 4 X 10^-4 = 0.0104
new [NH4Cl] = 0.0096/0.104 = 0.0923 M
[NH3] =0.0104/0.104 = 0.1 M
pOH = 4.745 + log 0.0923/0.1
pOH = 4.745 + log 0.923
pOH = 4.745 - 0.0347 = 4.7103
so new pH = 14 - 4.7103 = 9.2897
change = 9.2897 - 9.255 = 0.0347

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