Question

Calculate the change in pH when 5.00 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).

Answer 1

initial pH
henderson hasselbach formula
pH=pKa + log [base]/[acid]
[base]=[NH3]
[acid]=[NH4Cl]
pKa=pKw-pKb (Kw and Kb for water and NH3 are required)

Answer 2

mole HCl: n1=C1*V1=0.1*0.005=0.0005 moles
mole NH3:n2=C2*V2=0.1*0.1=0.01 moles
mole NH4Cl: n3=C3*V3=0.1*0.1=0.01 moles

HCl + NH3-> NH4Cl
initial n1 n2 n3
reacts/produses -n1 -n1 +n1
final 0 n2-n1 n3+n1

final [NH3]=0.0095/0.105
[NH4Cl]= 0.0105/0.105
final pH
pH=pKa + log [NH3]/[NH4Cl]
so
pH1=pKa
pH2=pKa + log 0.0095/0.105
ΔpH=log0.0095/0.105
ΔpH=- 1.043

Answer 3

please check my calc
i was not theral

Answer 4

There is a '0' in pH2 which should not be there.
Using millimoles instead of moles avoids working with too many 0's.
ΔpH= lg[9.5/10.5] = -0.04 which prooves the buffer effect of this solution.

Answer 5

@Vincent-Lyon.Fr your resalt makes more sence. althow i am looking to find the extra 0 i don't see it. not in a moode to do it from skrats. can you poit it out for me?

Answer 6

Sorry, it was not an extra 0, but a missing one.

This was ok :
final [NH3]=0.0095/0.105
[NH4Cl]= 0.0105/0.105
then pH2 should have read:
pH2=pKa + log 0.0095/0.0105 <--- here was the missing 0

I was confused because new volume 0.105 L and amount of NH4+ 0.0105 mol look very much the same.

Answer 7

yeap got it thanks man