A 1.0L buffer solution contains 0.100 mol of HC2H3O2

Question

and 0.100 mol of NaC2H3O2
The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 M HCl
to the original buffer.

Answer 1

I worked this buffer problem for someone yesterday, perhaps you, that asked for the pH after NaOH was added. I left instructions for the HCl addition. Now you want me to do it for you.
HC2H3O2 is HAc in shorthand notation and Ac^- is acetate ion.
pKa = -log Ka = 4.74
The Henderson-Hasselbalch equation is
pH = pKa + log (Ac^-)/(HAc)
You are adding 10 mL of 1 M HCl which is 0.01 x 1 = 0.01 mols.
At equilibrium you have
.........................Ac^- + H^+ ==> HAc
Initial mols........0.1......0.............0.1
add............................0.01..........................
change...........-0.01..-0.01.......... 0.01
equilibrium......0.09.......0..............0.11
Plug the equilibrium line into the HH equation and solve for pH.
Just a note here because some profs are picky. You will note that the HH equation uses CONCENTRATION OF Ac^- and HAc. I have used mols above. If you want to be proper, you can take the moles Ac and mols HAc and convert them to concn in mols/L. (Ac^-) = 0.09/1.01 L = ? and (HAc) = 0.115/1.01 L and use those instead of mols in the HH equation/ However, since the volume is the same in both cases (1.01 L), that number cancels. You get the same answer but picky profs will count off if moles are used instead of concn because concn is in the formula. Post your work if you get stuck. BTW, I was a picky prof and I gave only partial credit if moles were used without the volume. Some clever students who didn't want to go through the extra calculation to calculate M would show it as
pH = 4.74 + log (0.09/V)/(0.11/V). That I had to recognize that 0.09/V was the concentration. Those clever students received full credit and didn't go through the extra work either.