LONG ANSWER
Start by determining the concentrations of the two species in the buffer. For acetic acid, you have
#C = n/V => n_("acetic") = C * V = "0.2 M" * 100*10^(-3)"L" = 20 * 10^(-3)"moles"#
This means that the concentration of acetic acid in the buffer will be
#C_("acetic") = n_("acetic")/V_("buffer") = (20*10^(-3)"moles")/((100 + 100)*10^(_3)"L") = "0.1 M"#
For sodium acetate, the number of moles is
#n_("acetate") = C * V = "0.15 M" * 100 * 10^(-3)"L" = 15 * 10^(-3)"moles"#
As a result ,the concentration of sodium acetate in the buffer will be
#C_("acetate") = n_("acetate")/V_("buffer") = (15*10^(-3)"moles")/((100 + 100) * 10^(-3)"L") = "0.075 M"#
Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution
#pH_("solution") = pKa + log(([CH_3COO^(-)])/([CH_3COOH]))#
#pH_("solution") = 4.75 + log("0.075 M"/"0.1M") = 4.75 - 0.125 =4.63#
Now you add the #"HCl"# solution. The strong #"HCl"# acid will react with the wak sodium acetate base to produce weak acetic acid
#HCl_((aq)) + CH_3COONa_((aq)) -> NaCl_((aq)) + CH_3COOH_((aq))#
(I won't go into the net ionic equation)
The moles of hydrochloric acid added to the solution will be
#n_("HCl") = C * V = "0.2 M" * 20 * 10^(-3)"L" = 4 * 10^(-3)"moles"#
This means that the concentration of #"HCl"# in the buffer will be
#C_("HCl") = n_("HCl")/V_("buffer") = (4 * 10^(-3)"moles")/((100 + 100 + 20) * 10^(-3)"L") = "0.018 M"#
The new concetrations of acetic acid and sodium acetate will be
#C_("acetate") = (15 * 10^(-3)"moles")/(220 * 10^(-3)"L") = "0.068 M"#, and
#C_("acetic") = (20 * 10^(-3)"moles")/(220 * 10^(-3)"L") = "0.091 M"#
Now, all the hydrochloric acid will be consumed by the above reaction; this means that the concentration of sodium acetate will decrease by how much #"HCl"# was consumed, and the concentration of acetic acid will increase by the same amount.
Therefore,
#C_("acetic-final") = C_("acetic") + C_("HCl") = "0.091 M" + "0.018 M" = "0.109 M"#, and
#C_("acetate-final") = C_("acetate") - C_("HCl") = "0.068 M" - "0.018 M" = "0.05 M"#
This means that the solution's pH will now be
#pH_("sol") = 4.75 + log(("0.05 M")/("0.109 M")) = 4.75 - 0.338 = 4.41#
Notice how little the pH dropped despite the addition of a strong acid.
