A solution contains 0.09MHCl,0.09MCHCl2COOH, and 0.1MCH3COOH. If total [H+]=0.1 and Ka for CH3COOH=10−5,Ka for CCl2HCOOH is:
A
1.35×10−4
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B
0.18×10−2
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C
0.18×10−5
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D
1.25×10−2
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Solution
The correct option is C1.25×10−2 For a weak acid, the expression for the hydrogen ion concentration is [H+]=√c×Ka. Hence the total hydrogen ion concentraton of he solution will be, [H+]total=[H+]HCl+[H+]CHCl2COOH[H+]CH3COOH. [H+]total=0.1=0.09+√cCHCl2COOH×Ka,CHCl2COOH+√cCH3COOH×Ka,CH3COOH 0.01=√0.09×Ka,CHCl2COOH+√0.1×10−5 Hence, Ka,CHCl2COOH=9×10−4.