I have another chem question to solve help please!?

Question

this is dealing with a buffer lab we did in chem,

conc of acetic acid= 0.1 m, vol used= 20.00 ml

conc of CH3COONa= 0.1 m, vol used = 25.00 ml

conc of HCl= 0.1, vol used 5.00 ml

the ph that we measured using this electrode thing for the buffer was -> 5.1

and then the ph we measured after adding 5 ml of the HCl to the buffer solution was --> 5.0

now my assignment asks me to find the ph before the addition of hcl and also after the addition of hcl

i dont even know where to start! please help its due at 9 in the morning!

Answer 1

now my assignment asks me to

1) find the ph before the addition of hcl (you found pH = 5.1)

2) after the addition of hcl (you found pH = 5.0)

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1)  lets see what we can calculate the ph before the addition of hcl, to be theoretically

acetic acid was diluted 0.1 M (20 ml / 45ml) = 0.0444M

acetate was diluted 0.1 M (25ml/45 ml) = 0.0555M

acetic --> ---> H+   & acetate

0.0444M  -->  H+    &  0.0555M

K = [H+] [acetate] / [acetic]

1.8e-5 = [H+] [0.0555] / [0.0444]

H+ = 1.44e-5

your first answer: pH = 4.84

======================

2)  lets see what we can calculate the ph after the addition of hcl, to be theoretically

in adding 0.005litres of 0.1 mol/litre HCl, you added  0.0005 moles of H+, this shifts the equilibrium to the left:

acetic  <--- <--- H+   & acetate

making 0.0005 more moles of aectic acid & decreasing acetate by the same 0.0005 moles

acetic acid was 0.020 litres @ 0.1 mol/litre = 0.002

moles, it is now 0.0025 moles & it is in 50 ml.

0.0025 moles / 0.050 litres = 0.05 Molar

Acetate was 0/025 litres @ 0.1 mol/litre = 0.0025 moles, it is now 0.0020 moles & it is in 50 ml.

0.0020 mole / 0.050 litres = 0.04 molar

K = [H+] [acetate] / [acetic]

1.8e-5 = [H+] [0.04] / [0.05]

H+ = 2.25e-5

your next answer: pH = 4.65