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A solution is 0.1 Min formic acid Ka=2.0×104M. and 0.1 Min hydrofluoric acid Ka=8.0×104M. Its pH will be about

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By Expert Faculty of Sri Chaitanya
a
2.0
b
2.5
c
3.0
d
4.0

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detailed solution

Correct option is A

H3O+=K1c1+K2c2=2.0×104M(0.1M)+8.0×104M(0.1M)1/2 =1.0×102M pH=log1.0×102=2.0


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