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Calculate the pH of 0.00025 M HNO_(3). |
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Answer» Solution :`pH=-log_(10)[H^(+)]-log_(10)[0.00025]=-log_(10)[H^(+)]-log_(10)[2*5XX10^(-4)]=4-log2.5` = 4 - 0.3979 = 3.6021. |
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