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Question

The dissociation constant of a weak acid is 106. Then the PH of 0.01N of that acid is:

A
2
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B
7
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C
8
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D
4
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Solution

The correct option is C 4
Correct answer is D CH3COOHCH3COO+H+1kd=[CH3COO][H+1][CH3COO]kd=106Letybetheamountofaciddissociated(H+1ion)y2=0.01×106=108y=104pH=log(104)=4

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