Answer :
We start by determining the concentration of hydrogen ions in the hydrochloric acid solution from its pH. Recall that
[tex]$$
\text{pH} = -\log[H^+],
$$[/tex]
so the hydrogen ion concentration is given by
[tex]$$
[H^+] = 10^{-\text{pH}} = 10^{-3.5} \approx 3.16 \times 10^{-4} \text{ M}.
$$[/tex]
Next, we calculate the moles of [tex]$H^+$[/tex] present in the [tex]$50.0\ \text{mL}$[/tex] HCl solution. First, convert the volume to liters:
[tex]$$
50.0\ \text{mL} = 0.0500\ \text{L}.
$$[/tex]
Then, the moles of [tex]$H^+$[/tex] are:
[tex]$$
\text{moles of } H^+ = [H^+] \times \text{Volume} = 3.16 \times 10^{-4}\ \text{M} \times 0.0500\ \text{L} \approx 1.58 \times 10^{-5}\ \text{moles}.
$$[/tex]
The neutralization reaction between calcium hydroxide and hydrochloric acid is:
[tex]$$
\text{Ca(OH)}_2 + 2\ \text{HCl} \rightarrow \text{CaCl}_2 + 2\ \text{H}_2\text{O}.
$$[/tex]
This equation shows that one mole of [tex]$\text{Ca(OH)}_2$[/tex] reacts with two moles of [tex]$\text{HCl}$[/tex] (or equivalently, with two moles of [tex]$H^+$[/tex]). Therefore, the moles of [tex]$\text{Ca(OH)}_2$[/tex] required to neutralize the acid are half the moles of [tex]$H^+$[/tex]:
[tex]$$
\text{moles of } \text{Ca(OH)}_2 = \frac{1.58 \times 10^{-5}}{2} \approx 7.91 \times 10^{-6}\ \text{moles}.
$$[/tex]
Now, since this amount of [tex]$\text{Ca(OH)}_2$[/tex] is found in [tex]$200.0\ \text{mL}$[/tex] of solution, we convert the volume to liters:
[tex]$$
200.0\ \text{mL} = 0.2000\ \text{L}.
$$[/tex]
The molarity of the [tex]$\text{Ca(OH)}_2$[/tex] solution is then calculated as:
[tex]$$
M_{\text{Ca(OH)}_2} = \frac{\text{moles of } \text{Ca(OH)}_2}{\text{Volume in liters}} = \frac{7.91 \times 10^{-6}\ \text{moles}}{0.2000\ \text{L}} \approx 3.95 \times 10^{-5}\ \text{M}.
$$[/tex]
Thus, the molarity of the [tex]$\text{Ca(OH)}_2$[/tex] solution is
[tex]$$
\boxed{3.95 \times 10^{-5}\ \text{M}}.
$$[/tex]
[tex]$$
\text{pH} = -\log[H^+],
$$[/tex]
so the hydrogen ion concentration is given by
[tex]$$
[H^+] = 10^{-\text{pH}} = 10^{-3.5} \approx 3.16 \times 10^{-4} \text{ M}.
$$[/tex]
Next, we calculate the moles of [tex]$H^+$[/tex] present in the [tex]$50.0\ \text{mL}$[/tex] HCl solution. First, convert the volume to liters:
[tex]$$
50.0\ \text{mL} = 0.0500\ \text{L}.
$$[/tex]
Then, the moles of [tex]$H^+$[/tex] are:
[tex]$$
\text{moles of } H^+ = [H^+] \times \text{Volume} = 3.16 \times 10^{-4}\ \text{M} \times 0.0500\ \text{L} \approx 1.58 \times 10^{-5}\ \text{moles}.
$$[/tex]
The neutralization reaction between calcium hydroxide and hydrochloric acid is:
[tex]$$
\text{Ca(OH)}_2 + 2\ \text{HCl} \rightarrow \text{CaCl}_2 + 2\ \text{H}_2\text{O}.
$$[/tex]
This equation shows that one mole of [tex]$\text{Ca(OH)}_2$[/tex] reacts with two moles of [tex]$\text{HCl}$[/tex] (or equivalently, with two moles of [tex]$H^+$[/tex]). Therefore, the moles of [tex]$\text{Ca(OH)}_2$[/tex] required to neutralize the acid are half the moles of [tex]$H^+$[/tex]:
[tex]$$
\text{moles of } \text{Ca(OH)}_2 = \frac{1.58 \times 10^{-5}}{2} \approx 7.91 \times 10^{-6}\ \text{moles}.
$$[/tex]
Now, since this amount of [tex]$\text{Ca(OH)}_2$[/tex] is found in [tex]$200.0\ \text{mL}$[/tex] of solution, we convert the volume to liters:
[tex]$$
200.0\ \text{mL} = 0.2000\ \text{L}.
$$[/tex]
The molarity of the [tex]$\text{Ca(OH)}_2$[/tex] solution is then calculated as:
[tex]$$
M_{\text{Ca(OH)}_2} = \frac{\text{moles of } \text{Ca(OH)}_2}{\text{Volume in liters}} = \frac{7.91 \times 10^{-6}\ \text{moles}}{0.2000\ \text{L}} \approx 3.95 \times 10^{-5}\ \text{M}.
$$[/tex]
Thus, the molarity of the [tex]$\text{Ca(OH)}_2$[/tex] solution is
[tex]$$
\boxed{3.95 \times 10^{-5}\ \text{M}}.
$$[/tex]
