The dissociation constant for a weak monobasic acid is 4×10−10.pH for this 0.01N 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 : HA⇌H++A−
Applying Ostwald's dilution law:α=√KaC α=√4×10−100.01=2×10−4
Concentration of H+ [H+]=Cα=0.01×2×10−4[H+]=2×10−6molL−1pH=−log[H+]pH=−log[2×10−6]pH=5.7