A certain amount of #"BOH"# and #"BCl"# were mixed to make a #"200-mL"# buffer with #"pH" = 10.00#. If #"0.060 mols"# of #"HCl"# was added into the buffer and it results in a buffer #"pH"# of #9.00#, what was each concentration before adding acid?

I got #"0.3998 M BOH"# and #"0.2004 M BCl"#.

Well, sometimes you may not know what to do until you try it.

Obviously we are going to use the Henderson-Hasselbalch equation, because we are using a buffer. This is one version of it:

#"pH" = "pK"_a + log((["base"])/(["acid"]))#

At #25^@ "C"#, #"pK"_a + "pK"_b = "pK"_w = 14#, so #"pK"_a = 9.70# for #"BOH"_2^(+)#. As a result, we can calculate the ratio of base to acid:

#10^("pH" - "pK"_a) = (["base"])/(["acid"])#

#= 10^(10.00 - 9.70) = 1.995#

Therefore, #["base"] = 1.995["acid"]#.

Now, by adding #"HCl"# into the buffer, it reacts with the weak base #"BOH"# (consuming it) to generate more weak acid #"BCl"#.

Therefore, the new, lower #"pH"# was reached, giving this new ratio of base to acid that comes from the addition of strong acid:

#9.00 = 9.70 + log((["base"]cdot"0.200 L" - "0.060 mols HCl")/(["acid"]cdot"0.200 L" + "0.060 mols HCl"))#

We do not care about the final volume here, because it cancels out. So we use the initial buffer volume just to get the mols of weak acid and base in the log argument.

Using the ratio of INITIAL base concentration to acid concentration that we got previously (which has not changed), we can rewrite this in terms of one variable.

#9.00 = 9.70 + log((1.995["acid"]cdot"0.200 L" - "0.060 mols HCl")/(["acid"]cdot"0.200 L" + "0.060 mols HCl"))#

Solving for the log argument,

#10^(9.00 - 9.70) = 0.1995#

#= (1.995["acid"]cdot"0.200 L" - "0.060 mols HCl")/(["acid"]cdot"0.200 L" + "0.060 mols HCl")#

Now it's just algebra.

#0.1995["acid"]cdot"0.200 L" + 0.1995cdot"0.060 mols HCl" = 1.995["acid"]cdot"0.200 L" - "0.060 mols HCl"#

#0.1995cdot"0.060 mols HCl" + "0.060 mols HCl" = 1.7955["acid"]cdot"0.200 L"#

#1.1995cdot"0.060 mols HCl" = 1.7955["acid"]cdot"0.200 L"#

As a result:

#color(blue)(["acid"]) = (1.1995cdot"0.060 mols HCl")/(1.7955 cdot "0.200 L")#

#=# #color(blue)ul("0.2004 M BCl")#

Therefore:

#color(blue)(["base"]) = 1.995 cdot "0.2004 M BCl" = color(blue)ul("0.3998 M BOH")#

To check, do we still get a #"pH"# of #9.00#?

#"pH" = 9.70 + log(("0.3998 M BOH" cdot "0.200 L" - "0.060 mols HCl")/("0.2004 M BCl" cdot "0.200 L" + "0.060 mols HCl"))#

#= 9.70 + log(0.1994)#

#= 8.9998 ~~ 9.00# #color(blue)(sqrt"")#