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A solution has been prepared by dissolving 0.63 g of nitric acid in 100 mL. What is its pH value ? Assume that the acid is completely dissociated.

Answer» Molar concentration of the solution `=("Amount of"HNO_(3)//"litre of solution")/("Molar mass of"HNO_(3)) = (0.63)/(100) xx (1000)/(63)`
`=0.1 M`
the complete dissociation of acid in solution is :
`{:(HNO_(3)(l) ,+,H_(2)O(l) ,overset(aq)(to), H_(3)O^(+) ,+, NO_(3)^(-) (aq)),(,,,,0.1M,,0.1M):}`
`pH=-log [H_(3)O^(+)] =- log[10^(-1)] =(-) (-log 10) =1`


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