Question

Calculate the pH value of \[{\text{HN}}{{\text{O}}_3}\] solution containing \[0.315{\text{ g}}\] acid in \[{\text{200 ml}}\] of solution.
\[\left( {{\text{log2}}{\text{.5 = 0}}{\text{.3979 }}} \right)\]
A ) 1.6021
B ) 2
C ) 0.3979
D ) 1

Answer 1

Hint: pH of the solution is the negative logarithm of hydrogen ion concentration
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]\]
For nitric acid solution, the hydrogen ion concentration is equal to the nitric acid concentration. Molarity is the number of moles of solute present in one litre of solution.

Complete step by step answer:
The molar mass of nitric acid is \[63{\text{ g/mol}}\] . The mass of nitric acid is \[0.315{\text{ g}}\]. Divide mass of nitric acid with its molar mass to calculate the number of moles of nitric acid.
\[\dfrac{{0.315{\text{ g}}}}{{63{\text{ g/mol}}}} = {\text{ 0}}{\text{.005 mol}}\]
The volume of the solution is \[{\text{200 ml}}\]. Convert the unit of volume from millilitres to litres.
\[\dfrac{{200{\text{ mL}}}}{{1000{\text{ L/mL}}}} = {\text{ 0}}{\text{.200 L}}\]
Divide the number of moles of nitric acid with the volume of solution to obtain the molarity of nitric acid solution.
\[\dfrac{{0.005{\text{ mol}}}}{{0.200{\text{ L}}}} = {\text{ 0}}{\text{.025 M}}\]
From the concentration of nitric acid, calculate the concentration of hydrogen ions.
\[\left[ {{{\text{H}}^ + }} \right] = \left[ {{\text{HN}}{{\text{O}}_3}} \right] \\
\left[ {{{\text{H}}^ + }} \right] = {\text{ 0}}{\text{.025 M}} \\\]
Use the concentration of hydrogen ions to calculate the pH of the solution.
\[{\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}^ + }} \right] \\
{\text{pH}} = - {\log _{10}}{\text{0}}{\text{.025 M}} \\
{\text{pH}} = - \left( {{{\log }_{10}}\left( {{\text{1}}{{\text{0}}^{ - 2}}{\text{ M}} \times {\text{2}}{\text{.5 M}}} \right)} \right) \\\]
\[{\text{pH}} = - \left( {{{\log }_{10}}{\text{1}}{{\text{0}}^{ - 2}}{\text{ M}} + {{\log }_{10}}{\text{2}}{\text{.5 M}}} \right) \\
{\text{pH}} = - \left( { - 2 + 0.3979} \right) \\
{\text{pH}} = - \left( { - 1.6021} \right) \\
{\text{pH}} = 1.6021 \\\]
Thus, the pH of the solution is 1.6021.

So, the correct answer is “Option A”.

Note: Nitric acid is a strong acid. It completely dissociates into aqueous solution to provide hydrogen ions. Usually, the pH of strong acids is in the range 0 to 1.