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Question

A 0.2 molar solution of formic acid is 3.2% ionised, its ionisation constant is:

A
9.6×103
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B
2.1×104
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C
1.25×106
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D
2.1×108
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Solution

The correct option is D 2.1×104
Initial amount:At equillibrium:HCOOH0.2(0.2x)HCOO0x+H+0(x)

Ka=[HCOO][H+][HCOOH]=x×x(0.2x)

Given that 3.2% of 0.2 M formic acid is ionised.
x=3.2100×0.2=0.0064

Therefore, at equillibrium,
[HCOOH]=0.20.0064=0.1936
[HCOO]=0.0064
[H+]=0.0064

Ka=0.0064×0.00640.1936=0.4×1040.1936=2.1×104

Hence, option B is correct.

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