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

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

Step-by-step 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) * \frac{\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) *\frac{\text{2 mmol HNO}_(3)}{\text{1 mmol Ca(OH)}_(2)}= \text{4.642 mmol HNO}_(3)`

(d) Molar concentration of HNO₃

`c = \frac{\text{moles of solute}}{\text{litres of solution}}\\\\c = (n)/(V)\\\\c= \frac{\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 $\large \boxed{\textbf{0.195 mol/L}}$}`