I was wondering if you would be able to help me with a Chemistry question about the theoretical pH of a buffer solution. I have to calculate the theoretical pH values expected for a 200mL buffer solution containing a 1:1 ratio of acetic acid and sodium acetate (their concentrations are 0.05M), following the addition of;

Question

a) 10mL of 0.2M HCl and
b) 10mL of 0.2M NaOH.
Attempt for a:
[HCl] = 0.2M V = 10mL/0.01L thus n = 0.002
0.002 x 0.2L = 0.01M
........H3O + A <=> HA + H2O
I = x 0.05. 0.05
C = - 0.01. -0.01 +0.01
E = x-0.01. 0.04. 0.06
pH = 4.74 + log (0.06/0.04)
= 4.926
Although in our experiments the actual pH was 4.17 after 10mL of HCl was added and the pH after 10 mL of NaOH was 4.98
ATTEMPT for b
[NaOH] = 0.2M V = 10mL/0.01L thus n = 0.002
0.002 x 0.2L = 0.01M
-------HA + OH- <=> H2O + A-
I= 0.05. x 0.05
C= -0.01. x-0.01. + 0.01
E = 0.04. 0.06
pH = 4.74 + log(0.04/0.06)
pH = 3.74
So I feel like I have done these the wrong way because it makes more sense but I don't know why or how.

Answer 1

pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6, then, 7.21 will be used. This means, monopotassium dihydrogen phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.
pH = pKa + log ([A-]/[HA])
Now, for A- you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:
6 = 7.21 + log(A/HA)
log(A/HA)=-1.2
A/HA = 0.063
So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:
HA+A=0.05
A/HA=0.063
HA=0.047M
A=0.003M
This means, you need to put 0.047 moles of KH2PO4 and 0.003 moles of K2HPO4 salts into 1 liters of solution.
for molecular weights: KH2PO4= 39+97=136amu; K2HPO4= 39*2+96=174amu
Thus, 0.047*136=6.392g of KH2PO4 and 0.003*174= 0.522g of K2HPO4 should be added.
But, you should also note that, if |pH-pKa| >1 , then buffer capacity of solution decreases. In this case, it is equal to 1.2
I hope the calculations are clear.
Wish you luck.

Answer 2

I'm a bit confused with your calculations. We were only given one Ka which was for acetate, which I calculated the pKa and used it in the Henderson equation. I don't know where the KH2PO4 is coming from. I'm only meant to calculate it with the information given with this equilibrium
CH3COOH + H2O <---> CH3COONa + H3O
And my teacher said this should be solvable with the ICE analysis table easily. Just a bit confused can you please clear this up? Thank you