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- [Instructor] A
polyprotic acid is an acid
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with more than one proton that
it can donate in solution.
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An example of a polyprotic
acid is the protonated form
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of the amino acid alanine.
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Here's a dot structure
showing the protonated form
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of the amino acid alanine,
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and we can represent this as H2A plus.
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Let's say we're doing a titration
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with the protonated form of alanine,
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and we're adding sodium hydroxide
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to an aqueous solution
of the protonated form.
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The protonated form of alanine
has two acidic protons.
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So one acidic proton is on the oxygen,
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and the other acidic proton is one
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of the protons bonded to the nitrogen.
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The proton bonded to the oxygen
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is the more acidic of the two.
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Therefore, when we add
some hydroxide anions
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to the solution,
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the hydroxide anions will pick
up this proton and form water
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and convert the protonated form of alanine
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into the overall neutral
form of alanine, HA.
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Let's say that we start
out in our titration
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with one mole of the
protonated form of alanine,
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and then we add in our hydroxide anions.
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After we've added in one
mole of hydroxide anions,
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we've completely neutralized
the protonated form of alanine,
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and H2A plus is converted into HA.
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So one mole of H2A plus reacts
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with one mole of hydroxide anions
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to form one mole of HA.
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So now we have one mole of HA in solution.
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And if we continue to
add hydroxide anions,
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hydroxide anions react with HA
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and take one of the acidic
protons on the nitrogen
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to form A minus.
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A minus is the conjugate base to HA.
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It would take another
mole of hydroxide anions
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to completely neutralize the
HA and to turn it into A minus.
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So one mole of HA reacts with
one mole of hydroxide anions
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to form one mole of A minus.
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So going back to the protonated form
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of the amino acid alanine,
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since the protonated form
has two acidic protons,
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we call this a diprotic acid.
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So let me go ahead and
write in here diprotic.
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And because there are two acidic protons
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on the protonated form of alanine,
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there will be two pKa values,
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which we can figure out
from the titration curve.
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Next, let's look at the
titration curve for the titration
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of our diprotic acid
with sodium hydroxide.
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On the y-axis is pH,
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and on the x-axis is moles
of hydroxide anions added.
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Before we've added any hydroxide anion,
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so this point right here
on our titration curve,
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we have our diprotic acid present,
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so H2A plus.
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And since we started with
one mole of H2A plus,
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it would take one mole of hydroxide anions
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to completely neutralize the H2A plus.
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So after one mole of hydroxide
anions has been added,
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that brings us to equivalence point one.
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So we go to one mole of
hydroxide anions on the x-axis,
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and we'd go up to where that
intersects our titration curve.
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And this point on our titration curve
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is equivalence point one.
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At equivalence point one,
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all of the H2A plus has
been converted into HA.
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So now we have one mole of HA present.
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Next, let's think about adding 0.5 moles
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of hydroxide anions.
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So that would only neutralize half
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of the H2A plus that
was initially present.
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Therefore, if we go to 0.5
moles of hydroxide anions
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and we'd go up to where that
intersects our titration curve,
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this point represents the
half equivalence point
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or half equivalence point
one for this titration.
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Next, let's look at half
equivalence point one
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in more detail.
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So here is half equivalence point one
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on the titration curve.
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So if we started out
with one mole of H2A plus
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and we've added 0.5 moles
of hydroxide anions,
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we've neutralized half, or
0.5 moles, of the H2A plus.
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Therefore, at the half equivalence point,
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we have half a mole of H2A
plus and half a mole of HA,
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which means the concentration
of H2A plus is equal
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to the concentration of HA.
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And because at this point
we have significant amounts
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of both the weak acid
and its conjugate base,
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we've formed a buffer.
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So notice how the pH changes very slowly
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in this region around the
half equivalence point one.
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So this is called buffer
region one right in here.
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And since we've formed a buffer,
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the Henderson-Hasselbalch
equation tells us
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when the concentration
of the weak acid is equal
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to the concentration
of the conjugate base,
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the pH of the solution is equal
to the pKa of the weak acid.
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In this case, we're
talking about the pKa value
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of the more acidic proton,
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so the proton that's on
the oxygen, NH2A plus.
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So if we find the half equivalence point
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and we go over to where this
intersects with the y-axis,
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the pH at this point should
be equal to the pKa value
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for the more acidic proton.
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So looking at the y-axis here,
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it looks like it's a little bit over two.
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And in reality,
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the value of pKa one is equal to 2.34.
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Next, let's go back to the
first equivalence point,
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which occurred after we added
one mole of hydroxide anions.
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So at this point, we have HA.
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If we think about adding another
mole of hydroxide anions,
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so going from one mole
total to two moles total,
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the extra mole of hydroxide anions
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will completely neutralize the
HA and turn it into A minus.
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So one mole of HA reacts
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with one mole of hydroxide anions
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to form one mole of A minus.
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And since the extra
mole of hydroxide anions
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completely neutralize the
HA and turn it into A minus,
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after we've added a total of
two moles of hydroxide anions,
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we go up to where that
intersects our titration curve,
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and this point represents
equivalence point two.
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So the equivalence point one occurs
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after adding one mole of hydroxide anions,
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and we can see the pH by
going over to the y-axis.
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And equivalence point two
occurs after we've added a total
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of two moles of hydroxide anions.
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Let's go back to our
first equivalence point
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where we had one mole of HA.
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And at that point,
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we've already added one
mole of hydroxide anions.
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And so if we add another 0.5
moles of hydroxide anions
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to give a total of 1.5 moles,
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we would neutralize half of the HA
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and turn half of it into A minus.
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Therefore, we have another
half equivalence point
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after 1.5 moles of hydroxide
anions have been added.
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Let's look in more detail at
half equivalence point two,
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which is right here on our titration curve
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after we've added 1.5
moles of hydroxide anions.
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Since we've neutralized half of the HA
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that was present at equivalence point one,
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now the concentration of HA is equal
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to the concentration of A minus.
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And since we have significant amounts
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of both the weak acid
and its conjugate base,
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we have a buffer present in the region
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around half equivalence point two.
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So we can see the pH is changing
very slowly in this region
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around the half equivalence point
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as hydroxide anions are added.
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And so this represents buffer region two.
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And because the concentration
of weak acid is equal
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to the concentration
of its conjugate base,
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the Henderson-Hasselbalch
equation tells us the pH
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at this point is equal to the pKa value.
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In this case, it would be pKa two,
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so one of the acidic
protons on the nitrogen.
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So we can find the value for pKa two
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by locating half equivalence point two
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and going over to see where
that intersects with our y-axis.
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It looks to be a little bit under 10,
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which matches with the
actual pKa two value,
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which turns out to be 9.87.
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Finally, let's go back to
the two equivalence points
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for our titration curve.
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The number of equivalence
points in a titration curve
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for a polyprotic acid is equal
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to the number of acidic
protons in the acid.
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Therefore, since we
titrated a diprotic acid
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with two acidic protons,
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the titration curve has
two equivalence points.