millimola HAc (acetic acid) = mL x L = 100 mL x 0.064 M = 6.4 mmols.
millimols Ac^- (acetate ion) = mL x L = 100mL x 0.036 M = 3.6 mmols.
When HCl is added the acetate reacts to limit the pH.
...............Ac^- + H^+ --> HAc
I.............3.6........0.............6.4
add.....................1................................
C........... -1.........-1..............+1......................
E............2.6..........0.............7.4....................
Plug the E line into pH = pKa + log (Ac^-)/(HAc) and solve for pH.
Post your work if you get stuck.
what will be the pH of a 100mL buffer solution containing 0.064 M of acetic acid and 0.036M f sodium acetate, after addition of 1 millimole of the hydrochloric acid? pka=4.75
3 answers
ph=4.75 + log (2.6/7.4) = 4.30 that right?
Looks good to me.