To determine the molarity of the [tex]\( \text{Ba(OH)}_2 \)[/tex] solution, we can follow these steps:
1. Write down the balanced chemical equation:
[tex]\[
2 \text{HCl} + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2 \text{O} + \text{BaCl}_2
\][/tex]
2. Calculate the moles of HCl used in the titration:
- Given:
[tex]\[
V_{\text{HCl}} = 66.90 \, \text{mL}
\][/tex]
[tex]\[
M_{\text{HCl}} = 0.500 \, \text{M}
\][/tex]
- Convert the volume of HCl from mL to L:
[tex]\[
V_{\text{HCl}} = \frac{66.90 \, \text{mL}}{1000} = 0.06690 \, \text{L}
\][/tex]
- Calculate the moles of HCl:
[tex]\[
\text{moles}_{\text{HCl}} = M_{\text{HCl}} \times V_{\text{HCl}} = 0.500 \, \text{M} \times 0.06690 \, \text{L} = 0.03345 \, \text{moles}
\][/tex]
3. Use stoichiometry to find the moles of Ba(OH)_2:
- From the balanced chemical equation, we see that 2 moles of HCl react with 1 mole of Ba(OH)_2.
[tex]\[
2 \text{HCl} + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2 \text{O} + \text{BaCl}_2
\][/tex]
- The moles of Ba(OH)_2 can be calculated by dividing the moles of HCl by 2:
[tex]\[
\text{moles}_{\text{Ba(OH)}_2} = \frac{\text{moles}_{\text{HCl}}}{2} = \frac{0.03345 \, \text{moles}}{2} = 0.016725 \, \text{moles}
\][/tex]
4. Calculate the molarity of the Ba(OH)_2 solution:
- Given the volume of Ba(OH)_2 solution:
[tex]\[
V_{\text{Ba(OH)}_2} = 50.0 \, \text{mL}
\][/tex]
- Convert the volume of Ba(OH)_2 from mL to L:
[tex]\[
V_{\text{Ba(OH)}_2} = \frac{50.0 \, \text{mL}}{1000} = 0.0500 \, \text{L}
\][/tex]
- Calculate the molarity of Ba(OH)_2:
[tex]\[
M_{\text{Ba(OH)}_2} = \frac{\text{moles}_{\text{Ba(OH)}_2}}{V_{\text{Ba(OH)}_2}} = \frac{0.016725 \, \text{moles}}{0.0500 \, \text{L}} = 0.3345 \, \text{M}
\][/tex]
Therefore, the molarity of the [tex]\(\text{Ba(OH)}_2\)[/tex] solution is [tex]\( 0.3345 \, \text{M} \)[/tex].
