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.
What does mean by 1/5 and 4/5 stages please give a details explanation.
1 Answer
This seems to be saying
I got
#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
That means if
The first thing to do here is to find how much
As they are both the same concentration, whatever volume we use of
#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
#n_(1//5) = 1/5n_(NaOH)#
#n_(4//5) = 4/5n_(NaOH)#
1/5 OF THE WAY
At this point,
The
#"pH" = "pK"_a + log((["A"^(-)])/(["HA"]))#
We have that
#"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,
As before,
#"pH"_(4//5) = 3.745 + log(((4//5) "mols")/((1//5) "mols"))#
DIFFERENCE IN pH
So, the difference in
#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)#