Let's go through the solution step-by-step:
1. Convert volumes to liters:
- Volume of [tex]\( \text{HNO}_3 \)[/tex] in liters:
[tex]\[
22.5 \, \text{mL} = 22.5 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0225 \, \text{L}
\][/tex]
- Volume of [tex]\( \text{Ca(OH)}_2 \)[/tex] in liters:
[tex]\[
31.27 \, \text{mL} = 31.27 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.03127 \, \text{L}
\][/tex]
2. Calculate the moles of [tex]\( \text{Ca(OH)}_2 \)[/tex]:
- Molarity of [tex]\( \text{Ca(OH)}_2 \)[/tex] is given as [tex]\( 0.167 \, \text{M} \)[/tex]:
[tex]\[
\text{moles of Ca(OH)}_2 = \text{Molarity} \times \text{Volume in liters}
\][/tex]
[tex]\[
\text{moles of Ca(OH)}_2 = 0.167 \, \text{M} \times 0.03127 \, \text{L} = 0.00522209 \, \text{mol}
\][/tex]
3. Using the balanced chemical equation to find moles of [tex]\( \text{HNO}_3 \)[/tex]:
- The balanced chemical equation is:
[tex]\[
2 \text{HNO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Ca(NO}_3\text{)}_2 + 2 \text{H}_2\text{O}
\][/tex]
- This equation shows that 2 moles of [tex]\( \text{HNO}_3 \)[/tex] react with 1 mole of [tex]\( \text{Ca(OH)}_2 \)[/tex]. Therefore, the mole ratio of [tex]\( \text{HNO}_3 \)[/tex] to [tex]\( \text{Ca(OH)}_2 \)[/tex] is 2:1.
[tex]\[
\text{moles of HNO}_3 = 2 \times \text{moles of Ca(OH)}_2
\][/tex]
[tex]\[
\text{moles of HNO}_3 = 2 \times 0.00522209 \, \text{mol} = 0.01044418 \, \text{mol}
\][/tex]
4. Calculate the molarity of [tex]\( \text{HNO}_3 \)[/tex]:
- Molarity is defined as the number of moles of solute per liter of solution:
[tex]\[
\text{Molarity of HNO}_3 = \frac{\text{moles of HNO}_3}{\text{volume of HNO}_3 \text{ in liters}}
\][/tex]
[tex]\[
\text{Molarity of HNO}_3 = \frac{0.01044418 \, \text{mol}}{0.0225 \, \text{L}} = 0.4641857777777778 \, \text{M}
\][/tex]
Therefore, the molarity of the [tex]\( \text{HNO}_3 \)[/tex] solution is approximately [tex]\( 0.464 \, \text{M} \)[/tex].
