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Question: Calculate the pH value of \[{\text{HN}}{{\text{O}}_3}\] solution containing \[0.315{\text{ g}}\] aci...

Calculate the pH value of HNO3{\text{HN}}{{\text{O}}_3} solution containing 0.315 g0.315{\text{ g}} acid in 200 ml{\text{200 ml}} of solution.
(log2.5 = 0.3979 )\left( {{\text{log2}}{\text{.5 = 0}}{\text{.3979 }}} \right)
A ) 1.6021
B ) 2
C ) 0.3979
D ) 1

Explanation

Solution

pH of the solution is the negative logarithm of hydrogen ion concentration
pH=log10[H+]{\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 g/mol63{\text{ g/mol}} . The mass of nitric acid is 0.315 g0.315{\text{ g}}. Divide mass of nitric acid with its molar mass to calculate the number of moles of nitric acid.
0.315 g63 g/mol= 0.005 mol\dfrac{{0.315{\text{ g}}}}{{63{\text{ g/mol}}}} = {\text{ 0}}{\text{.005 mol}}
The volume of the solution is 200 ml{\text{200 ml}}. Convert the unit of volume from millilitres to litres.
200 mL1000 L/mL= 0.200 L\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.
0.005 mol0.200 L= 0.025 M\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] = {\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.