Answer :
Answer:
[tex]\large \boxed{\text{0.195 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O
(b) Moles of Ca(OH)₂
[tex]\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}[/tex]
(c) Moles of HNO₃
The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂
[tex]\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}[/tex]
(d) Molar concentration of HNO₃
[tex]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 $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]