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Question

25 mL of 0.1 N H3PO4 is titrated against 0.1 N NaOH. The pH of the solution after addition of 62.5 mL of the base is :
(Given : Ka1=1×103,Ka2=1×108 and Ka3=1×1013)

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Solution

H3PO4+NaOHNaH2PO4

NaH2PO4+NaOHNa2HPO4
Na2HPO4+NaOHNa3PO4
First equivalence point is reached after addition of 25 mL of the base.
Second equivalence point is reached after addition of 50 ml of the base.
After addition of 62.5 mL all the H3PO4 and NaH2PO4 are consumed and the half equivalence point of third tiration is reached.

Species present in the solution at this point will be,
Na2HPO4 and Na3PO4 and its a buffer

pH=pKa3+log[Na3PO4][Na2HPO4]
since [Na2HPO4]=[Na3PO4]
pH=pKa3
Ka3=1×1013
Hence, pH=13


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