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Question

Carbonic acid (H2CO3), a diprotic acid has Ka1=4.0×107 and Ka2=7.0×1011. What is the [HCO3] of a 0.025 M solution of carbonic acid?

A
7.8×103
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B
6.6×104
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C
1010
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D
1.0×104
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Solution

The correct option is D 1.0×104
H2CO3HCO3+H+ka1
HCO3CO23+H+Ka2
c=0.0025H
Most will be from Ka,
α=Ka1C
4×10725×103
4×104α5
=25×102
=0.025=0.004<0.1
[H+]=cα
=0.0025×0.004
=25×4×106
=104
[h+][HCO3][H2CO3]=Ka,
=104×[HCO3]0.025=4×107
[HCO3]=4×103×25×103
=104

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