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Question

A solution contains 0.09 M HCl,0.09 M CCl2HCOOH, and 0.1 M CH3COOH. If total [H+]=0.1 and Ka for CH3COOH=105. Ka for CCl2HCOOH is:

A
1.35×104
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B
0.18×102
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C
0.18×105
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D
0.9×104
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Solution

The correct option is D 0.9×104
For a weak acid, the expression for the hydrogen ion concentration is
[H+]=e×Ka
Hence the total hydrogen ion concentration of the solution will be,
[H+]total=[H+]HCl+[H+]CHCl2COOH[H+]CH3COOH
[H+]total=0.1=0.09+CCHCl2COOH×KaCHCl2COOH+CCH3COOH×KaCH3COOH
0.01=0.09×KaCHCl2COOH+0.1×105
Hence, Ka, CHCl2COOH=0.9×104

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