the pH of 0.16 m
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Normality= molarity ×acidity
N=0.16×1
N=OH concentration
PoH=-log[OH]=log 16-log0.01=1.2+2=3.2
PH=14-PoH
PH=14-3.2=10.8
Therefore, your answer is 10.8
N=0.16×1
N=OH concentration
PoH=-log[OH]=log 16-log0.01=1.2+2=3.2
PH=14-PoH
PH=14-3.2=10.8
Therefore, your answer is 10.8
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