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Question

0.1 mole arsenic acid (H3AsO4) is dissolved in a 1L buffer solution of pH=8. Which of the following options is/are correct? For arsenic acid: Ka1=2.5×104,Ka2=5×108,Ka3=2×1013.
(‘<<’ sign denotes that the higher concentration is 100 times greater than the lower one or more)

A
[H3AsO4]<<[H2AsO4]
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B
[H2AsO4]<<[HAsO24]
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C
[HAsO24]<<[H2AsO4]
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D
[AsO34]<<[HAsO24]
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Solution

The correct option is D [AsO34]<<[HAsO24]
Since H3AsO4 is a polybasic acid and given that[H+]=108
it will ionise like
H3AsO4H2AsO4+H+
Ka1=[H2AsO4][H+][H3AsO4]
[H3AsO4][H2AsO4]=[H+]Ka1=1082.5×104
[H3AsO4]<<[H2AsO4]
similarly for second ionisation
[HAsO24][H2AsO4]=[H+]Ka2=1085×108=15
for third ionisation
[HAsO24][AsO34]=[H+]Ka3=1082×1013=5×104
Hence
[AsO34]<<[HAsO24]


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