Question

A 13.1 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 22.8 mL of 0.456 M potassium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture? % by mass?

Answer 1

Consider the neutralization reaction,

`HNO_3(aq) + KOH(aq) to KNO_3(aq) + H_2O(l)`

If it takes,

`0.456M * 0.0228L approx 0.0104"mol"`

of hydroxide ions to fully neutralize the sample, then there must be an equivalent amount of nitric acid molecules.

Hence,

`0.0104"mol" * (63"g")/(HNO_3) approx (0.655"g")/(13.1"g") * 100% approx 5.00%`

is the percent by mass of the original solution of nitric acid.