0.1M Formic Acid solution is titrated against 0.1 M NaOH solution.What would be the difference in pH between 1/5 and 4/5 stages of neutralization of acid?

What does mean by 1/5 and 4/5 stages please give a details explanation.

1 Answer
Truong-Son N.
Feb 14, 2018

This seems to be saying #1/5# and #4/5# of the way to the equivalence point, so if we can find the equivalence point, we can find the volumes needed of #"NaOH"#, and therefore solve each resultant equilibrium.

I got #Delta"pH"_(1//5 -> 4//5) = 1.20#. To check this,

#1/5 xx 100% = 20%# of the way to the equivalence point.

#4/5 xx 100% = 80%# of the way to the equivalence point.

#2.5/5 xx 100% = 50%# is the half-equivalence point, where #"pH" = "pK"_a#.

There is an odd function symmetry about the half-equivalence and equivalence points, so we should be equidistant from the #"pK"_a# in either direction.

That means if #"pH"_(1//2) - "pH"_(1//5) ne "pH"_(4//5) - "pH"_(1//2)#, you missed something.

The first thing to do here is to find how much #"NaOH"# was needed to fully neutralize #"0.1 M"# #"CHOOH"#, which we'll just call #"HA"# for simplicity as it is a monoprotic acid.

As they are both the same concentration, whatever volume we use of #"NaOH"# will be equal to the volume of formic acid. That means #1/5# and #4/5# of the way to the equivalence point,

#V_(1//5) = 1/5V_(NaOH)#

#V_(4//5) = 4/5V_(NaOH)#

If we have a fraction of a volume, we have a fraction of the starting mols at the given concentration. We know that the same number of mols react between the formic acid and #"NaOH"#, so we can treat this as:

#n_(1//5) = 1/5n_(NaOH)#

#n_(4//5) = 4/5n_(NaOH)#

1/5 OF THE WAY

At this point, #4/5# of the #"mols"# of #"OH"^(-)# are left, and #1/5# of the #"mols"# of #"OH"^(-)# neutralized #1/5# of the #"mols"# of #"HA"# to give the #"mols"# of #"A"^(-)#. We now have both #"A"^(-)# and #"HA"# in solution, so we are in the buffer region.

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The #K_a# of formic acid is #1.8 xx 10^(-4)#, which is fairly small, and at this concentration would give less than #5%# dissociation. Therefore, the Henderson-Hasselbalch equation applies:

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

We have that #"pK"_a = -log(K_a) = 3.745#, so:

#"pH"_(1//5) = 3.745 + log(((1//5) "mols")/((4//5) "mols"))#

and we can use mols since the total volume is shared. Also, since the total mols of each reactant are equal, they will cancel to give only unitless fractions.

4/5 OF THE WAY

At this point, #1/5# of the #"mols"# of #"OH"^(-)# are left, and #4/5# of the #"mols"# of #"OH"^(-)# neutralized #4/5# of the #"mols"# of #"HA"# to give the #"mols"# of #"A"^(-)#. We now still both #"A"^(-)# and #"HA"# in solution, so we are still in the buffer region.

As before,

#"pH"_(4//5) = 3.745 + log(((4//5) "mols")/((1//5) "mols"))#

DIFFERENCE IN pH

So, the difference in #"pH"# going from #1/5# to #4/5# of the way to the equivalence point is:

#color(blue)(Delta"pH"_(1//5 -> 4//5)) = "pH"_(4//5) - "pH"_(1//5)#

#= [cancel(3.745) + log(((4//5) "mols")/((1//5) "mols"))] - [cancel(3.745) + log(((1//5) "mols")/((4//5) "mols"))]#

#= log(((4//5) "mols")/((1//5) "mols")) - log(((1//5) "mols")/((4//5) "mols"))#

#= log(((4//5) "mols")/((1//5) "mols")) + log(((4//5) "mols")/((1//5) "mols"))#

#= 2log(((4//5) "mols")/((1//5) "mols"))#

#= 2log(4)#

#= color(blue)(1.20)#

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