1.

Calculate the pH of 0.00025 M HNO_(3).

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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