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.

Explanation

Concentration of `HNO_(3)` solution = 0.63 g per 100 ml (Given)
`=6.3 g ` per litre `= (6.3)/(63) ` moles/litre = `10^(-1)M` (`:' ` Mol. Mass of `HNO_(3) = 63`)
Now, `HNO_(3)`, completely ionizes as : `HNO_(3)H_(2)O rarrH_(3)O^(+)+NO_(3)^(-)`
`:. [H_(3)O^(+)]=[HNO_(3)]=10^(-1)M :. pH = - log [H_(3)O^(+)]= - log 10^(-1)=1`