Question

The
titration of 0.02500 L of a diprotic acid solution with 0.1000 M
NaOH requires 34.72 mL of titrant to reach the second equivalence
point. The pH is 3.95 at the first equivalence point and 9.27 at
the second equivalence point.

What is the pKa1 and pKa2 of the acid?

Thank You

Answer #1

Lets consider dissociation of a diprotic acid H_{2}X

H_{2}X <---------> HX^{-} +
H^{+}

K_{a}1 = [HX^{-} ][H^{+}] /
[H_{2}X]

At half-equivalence point half of H_{2}X is
neutralised,

Hence [H_{2}X] = [HX^{-} ]

So K_{a}1 = [H^{+}]

Or LogK_{a}1 = Log[H^{+}]

pKa1 = pH

Similarly it can be shown pKa2 = pH_{2}

pH_{2} = pH at second half-equivalence point

**Therefore pKa of an acid is equal to the pH a**t
**half-equivalence
point.**

In our problem pH at first half equivalence point = 3.95

Hence pH at first half-equivalence point = 3.95 / 2 = 1.975

Hence pKa1 = 1.975

Similarly pKa2 = 9.27 / 2 = 4.635

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Hint: For this type of problem, first calculate the volume of
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1)calculate the volume of titrant required to reach the second
equivalence point,
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3)calculate the pH after the addition of 62.0 mL of titrant,
4)calculate the pH after the addition of 75.0 mL of titrant.
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