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Question

Ka for the reaction; Fe3+(aq.)+H2O(l)Fe(OH)2+(aq.)+H3O+(aq.) is 6.5×103. What is the maximum pH value which could be used do that at least 80% of the total ion (III) in a dilute solution exists as Fe3+?

A
2
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B
2.41
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C
2.79
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D
0.59
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Solution

The correct option is D 0.59
Fe3+(aq.)+H2O(l)Fe(OH)2+(aq.)+H3O+(aq.)
At eqm 0.8c 0.2c 0.2c

Ka=[Fe(OH)2+][H3O+][Fe3+]=c×0.220.8=6.5×103;c=0.13M.

So pH=logc=log(0.2×0.13)=0.59.

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