What volume of 0.08892 M HNO₃ is required to react completely with 0.2352 g of potassium hydrogen phosphate?

\(2{\rm{HN}}{{\rm{O}}_{3({\rm{aq}})}}{\rm{ + }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_{4({\rm{aq)}}}}{\rm{ + 2KN}}{{\rm{O}}_{3({\rm{aq}})}}\)

The number of moles is the ratio of the given mass and molar mass of the compound. As the given mass of \({{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_4}\)is 0.2352g and molecular mass of \({{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_4}\)is \(39{\rm{ }} \times {\rm{ 2 + 1 + 30 + 16 }} \times {\rm{ }}4 = {\rm{ 173 g/mol}}\)

Therefore, the number of moles = \(\frac{{0.2352}}{{173}}{\rm{ = 1}}{\rm{.35 }} \times {\rm{ }}{10^{ - 3}}{\rm{ moles}}\)

Now, as 2 moles of nitric acid is required to react with one mole of \({{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_4}\), thus the number of moles of nitric acid will be \({\rm{1}}{\rm{.35 }} \times {\rm{ }}{10^{ - 3}}{\rm{ }} \times {\rm{ 2 = 2}}{\rm{.70 }} \times {\rm{ }}{10^{ - 3}}{\rm{ moles}}\)

The volume of nitric acid according to the mole concept can be calculated as:

\({\rm{Volume = }}\frac{{{\rm{Moles}}}}{{{\rm{Molarity}}}}\)

\({\rm{Volume = }}\frac{{{\rm{2}}{\rm{.70 }} \times {\rm{ }}{{10}^{ - 3}}}}{{0.08892}}{\rm{ = 0}}{\rm{.0304 L}}\)