Question

An aqueous solution of nitric acid is standardized by titration with a 0.110 M solution of calcium hydroxide. If 21.1 mL of base are required to neutralize 23.8 mL of the acid, what is the molarity of the nitric acid solution? ? M Nitric acid

Answer 1

Answer:

$\large \boxed{\text{0.195 mol/L}}$

Explanation:

(a) Balanced equation

2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O  

(b) Moles of Ca(OH)₂

$\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}$

(c) Moles of HNO₃

The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂

$\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}$

(d) Molar concentration of HNO₃

$c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol \cdot L ^{-1} }\\\\\text{The molar concentration of the Ca(OH) _{2} is \large \boxed{\textbf{0.195 mol/L}} }$