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Question

The dissociation constant for a weak monobasic acid is 4×1010. pH for this 0.01 N weak monobasic acid is :

A
9.7
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B
1.7
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C
3.7
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D
5.7
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Solution

The correct option is D 5.7
Let, the monobasic acid is HA. Its dissociation is given below :
HAH++A
Applying Ostwald's dilution law:α=KaC
α=4×10100.01=2×104
Concentration of H+
[H+]=Cα=0.01×2×104[H+]=2×106 mol L1pH=log[H+]pH=log[2×106]pH=5.7

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