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PH of 0.05 monobasic acid is 6.2. The % dissociation of it is?A) 1.29B) 0.029C) 2.12D) 0.04Please answer |
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Answer» Explanation: ph=6.2 so, Ka=10^-6.2 .......... {^}denotes = power Now, DISSOCIATION =√Ka/concentration =√10^-6.2/0.05 approximately =0.004 so, %=0.004*100 =0.4 This is CORRECT answer. please recheck your OPTIONS. |
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