Question

A chemist dissolves 751.mg of pure nitric acid in enough water to make up 290.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

Answer 1

1.4

Step-by-step explanation:

Mass of pure nitric acid = 751mg

Volume of solution = 290mL

Unknown:

pH of the solution = ?

Solution:

To solve this problem, we need the concentration of the acid in the aqueous form.

This is given by molarity;

Molarity =
`(number of moles )/(volume)`

Since the number of moles of nitric acid is unknown, we can easily solve for it.

Number of moles of nitric acid =
`(mass)/(molar mass)`

molar mass of HNO₃ = 1 + 14 + 3(16) = 63g/mol

mass of nitric acid = 751mg = 0.751g

Number of moles =
`(0.751)/(63)` = 0.012mole

Volume of solution = 290mL = 0.29dm³

Now molarity of the solution =
`(0.012)/(0.29)` = 0.041moldm⁻³

Since:

pH = -log [H₃O⁺]

HNO₃ + H₂O → H₃O⁺ + NO₃⁻

1moldm⁻³ 1moldm⁻³ 1moldm⁻³

0.041moldm⁻³ 0.041moldm⁻³ 0.041moldm⁻³

pH = -log[0.041] = 1.4