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
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)