Using Henderson-Hasselbalch equation, calculate the volume of 0.20 M Acetic Acid and 0.2 M Sodium Acetate needed to prepare 50 mL of 0.1 M Acetate Buffer solution (pH = 4.5). pKa of Acetic Acid is 4.74?

This can be contradictory, depending on whether the #"0.1 M"# is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

#V_(A^(-)) = "9.125 mL"#

#V_(HA) = "15.875 mL"#

The Henderson-Hasselbalch equation is:

#"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])#

We have a #"pH 4.5"# solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

#\frac(["A"^(-)])(["HA"]) = 10^("pH" - "pK"_a)#

#= 10^(4.5 - 4.74)#

#= 0.5754#

Now, if the total concentration is #"0.10 M"#, then:

#["HA"] + overbrace(0.5754["HA"])^(["A"^(-)]) = "0.10 M"#

#=> ["HA"] = ("0.10 M")/(1.0000 + 0.5754) = ul"0.0635 M"#

#=> ["A"^(-)] = ul"0.0365 M"#

and these concentrations are AFTER mixing. Since the total volume is #"50 mL"#, or #"0.050 L"#, the mols of each component (which are constant!) are:

#n_(A^-) = "0.0365 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.001825 mols"#

#n_(HA) = "0.0635 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.003175 mols"#

So, if both of the starting concentrations were #"0.20 M"#, we can find the volume they each start with:

#color(blue)(V_(A^-)) = "1 L"/(0.20 cancel("mols A"^(-))) xx 0.001825 cancel("mols A"^(-))#

#= "0.009125 L" = color(blue)ul("9.125 mL")#

#color(blue)(V_(HA)) = "1 L"/(0.20 cancel("mols HA")) xx 0.003175 cancel("mols HA")#

#= "0.015875 L" = color(blue)ul("15.875 mL")#

And this should make sense, because the total starting volume is #"25.000 mL"#, the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.