Calculate the change in pH for a buffer solution with an initial pH of #3.9# after addition of #"0.1 M HCl"#?

I think you might be asking, "calculate the change in #"pH"# for a buffer solution of #"pH"_i = 3.9# after addition of #"0.1 M"# #"HCl"#"...

Well, any properly made buffer would have the weak acid #"HA"# and its conjugate base #"A"^(-)#, or the weak base #"B"# and its conjugate acid #"BH"^(+)#. Added #"HCl"# will react with the base.

It should go without saying that any concentration without the volume added is ambiguous... Furthermore, you did not tell us what species are in solution, and in what concentrations...

So I will include three cases of an arbitrary acid and base at arbitrary concentrations... too bad for you I suppose?

1. Some acid added, but not enough to get to half-equivalence point.
2. Enough acid added to get to half equivalence point.
3. Enough acid added to get to equivalence point.

As always, we use the Henderson-Hasselbalch equation when not yet at the equivalence point.

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

Let's use formic acid for kicks...

#3.9 = 3.75 + log \frac(["A"^(-)])(["HA"])#

Currently we thus have

#\frac(["A"^(-)])(["HA"]) = 10^(3.9 - 3.75) = 1.413#, mostly weak base...

Let's suppose we therefore have a #"1-L"# solution of #"0.007063 M A"^(-)# and #"0.005000 M HA"#.

CASE I: NOT YET AT HALF EQUIV PT

We are still in the buffer region of the titration curve.

Adding, arbitrarily, #"5 mL"# of #"HCl"# gives...

#color(blue)("pH"') = 3.75 + log (("0.007063 mols" - "0.1 M" xx "0.005 L")/("0.005000 mol" + "0.1 M" xx "0.005 L"))#

#= color(blue)(ul(3.827))#,

as the mols of #"HCl"#, denoted #n_(HCl)#, neutralize some mols of the conjugate base, #n_(A^(-))#, to generate equimolar quantities of the weak acid, #n_(HA)#.

And we have not yet passed the half-equivalence point.

CASE II: AT HALF EQUIV PT

This case is easy; at the half-equivalence point, #["HA"] = ["A"^(-)]#, and thus #"pH" = "pKa" = 3.75#.

#color(blue)("pH"') = 3.75 + log((1)/(1)) = color(blue)(ul(3.75))#

So we might as well take it up a notch and calculate the volume of #"HCl"# that results in this situation.

#("0.007063 mols" - n_(HCl))/("0.005000 mols" + n_(HCl)) = 1#

#= 0.007063 - n_(HCl) = 0.005000 + n_(HCl)#

#=> n_(HCl) = 0.002063/2 = "0.001031 mols"#

And thus, the volume of #"HCl"# needed to get here is:

#V_(HCl) = "1000 mL"/(0.1 cancel"mols HCl") xx 0.001031 cancel"mols" = ul"10.314 mL"#

CASE III: AT EQUIV PT

Here, we've completely neutralized all of the weak base and only have weak acid remaining.

Thus, the Henderson-Hasselbalch equation does NOT apply yet, and we can only write the dissociation reaction for now:

#"HA"(aq) + "H"_2"O"(l) rightleftharpoons "A"^(-)(aq) + "H"_3"O"^(+)(aq)#

with #"pK"_a = 3.75# and #K_a = 1.778 xx 10^(-4)#.

At the start of this dissociation, we have a certain amount of #"HA"# generated from #"HCl"# reacting with #"A"^(-)# to add to the original concentration first:

#["HA"]_i = (overbrace("0.005000 mols")^("mols HA") + overbrace("0.1 M" xx "0.07063 L")^"mols HCl added")/(underbrace("1 L")_"Soln volume" + underbrace("0.07063 L")_("HCl added")#

#=# #"0.01127 M"#

Then, the mass action expression is:

#K_a = 1.778 xx 10^(-4) = (["A"^(-)]["H"_3"O"^(+)])/(["HA"]) = x^2/(0.01127 - x)#

And from solving the quadratic equation for

#1.778 xx 10^(-4)(0.01127) - 1.778 xx 10^(-4)x - x^2 = 0#,

we would get the physical answer, #x = "0.001329 M"# for the concentration of #"A"^(-)# after the dissociation of #"HA"#.

This means the current acid to base ratio is:

#(["A"^(-)])/(["HA"]) = ("0.001329 M")/("0.01127 M" - "0.001329 M") = 0.1337#

And NOW, we can calculate the #"pH"# using the Henderson-Hasselbalch equation, since we have a valid ratio that does not tend to infinity.

We are currently to the right of the equivalence point.

#color(blue)("pH"') = 3.75 + log(0.1337)#

#= color(blue)(ul(2.876))#