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Answer :
We begin by converting all volumes to liters. The 25.0 mL sample of saturated [tex]\( Ca(OH)_2 \)[/tex] is
[tex]$$
V_{\text{sample}} = \frac{25.0 \text{ mL}}{1000} = 0.0250 \text{ L},
$$[/tex]
and the volume of HCl at the equivalence point is
[tex]$$
V_{\text{HCl}} = \frac{35.9 \text{ mL}}{1000} = 0.0359 \text{ L}.
$$[/tex]
Since the titration uses 0.026 M HCl, the number of moles of HCl used is given by
[tex]$$
n_{\text{HCl}} = M \times V = 0.026 \, \text{M} \times 0.0359 \, \text{L} = 0.0009334 \, \text{mol}.
$$[/tex]
The balanced chemical equation for the reaction is
[tex]$$
Ca(OH)_2 + 2\,HCl \rightarrow CaCl_2 + 2\,H_2O.
$$[/tex]
This shows that one mole of [tex]\( Ca(OH)_2 \)[/tex] reacts with 2 moles of HCl. Therefore, the moles of [tex]\( Ca(OH)_2 \)[/tex] present in the sample are
[tex]$$
n_{Ca(OH)_2} = \frac{n_{\text{HCl}}}{2} = \frac{0.0009334}{2} = 0.0004667 \, \text{mol}.
$$[/tex]
When [tex]\( Ca(OH)_2 \)[/tex] dissociates, it produces two moles of hydroxide ions for every mole of [tex]\( Ca(OH)_2 \)[/tex]. Thus, the moles of [tex]\( OH^- \)[/tex] ions produced are
[tex]$$
n_{OH^-} = n_{Ca(OH)_2} \times 2 = 0.0004667 \times 2 = 0.0009334 \, \text{mol}.
$$[/tex]
Finally, the concentration of hydroxide ions in the solution is calculated by dividing the number of moles of [tex]\( OH^- \)[/tex] by the sample volume:
[tex]$$
[\text{OH}^-] = \frac{n_{OH^-}}{V_{\text{sample}}} = \frac{0.0009334 \, \text{mol}}{0.0250 \, \text{L}} = 0.037336 \, \text{M}.
$$[/tex]
Thus, the concentration of the hydroxide ion is approximately
[tex]$$
\boxed{0.037336 \, \text{M}}.
$$[/tex]
[tex]$$
V_{\text{sample}} = \frac{25.0 \text{ mL}}{1000} = 0.0250 \text{ L},
$$[/tex]
and the volume of HCl at the equivalence point is
[tex]$$
V_{\text{HCl}} = \frac{35.9 \text{ mL}}{1000} = 0.0359 \text{ L}.
$$[/tex]
Since the titration uses 0.026 M HCl, the number of moles of HCl used is given by
[tex]$$
n_{\text{HCl}} = M \times V = 0.026 \, \text{M} \times 0.0359 \, \text{L} = 0.0009334 \, \text{mol}.
$$[/tex]
The balanced chemical equation for the reaction is
[tex]$$
Ca(OH)_2 + 2\,HCl \rightarrow CaCl_2 + 2\,H_2O.
$$[/tex]
This shows that one mole of [tex]\( Ca(OH)_2 \)[/tex] reacts with 2 moles of HCl. Therefore, the moles of [tex]\( Ca(OH)_2 \)[/tex] present in the sample are
[tex]$$
n_{Ca(OH)_2} = \frac{n_{\text{HCl}}}{2} = \frac{0.0009334}{2} = 0.0004667 \, \text{mol}.
$$[/tex]
When [tex]\( Ca(OH)_2 \)[/tex] dissociates, it produces two moles of hydroxide ions for every mole of [tex]\( Ca(OH)_2 \)[/tex]. Thus, the moles of [tex]\( OH^- \)[/tex] ions produced are
[tex]$$
n_{OH^-} = n_{Ca(OH)_2} \times 2 = 0.0004667 \times 2 = 0.0009334 \, \text{mol}.
$$[/tex]
Finally, the concentration of hydroxide ions in the solution is calculated by dividing the number of moles of [tex]\( OH^- \)[/tex] by the sample volume:
[tex]$$
[\text{OH}^-] = \frac{n_{OH^-}}{V_{\text{sample}}} = \frac{0.0009334 \, \text{mol}}{0.0250 \, \text{L}} = 0.037336 \, \text{M}.
$$[/tex]
Thus, the concentration of the hydroxide ion is approximately
[tex]$$
\boxed{0.037336 \, \text{M}}.
$$[/tex]
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