Thank you for visiting A 25 0 mL sample of a saturated tex Ca OH 2 tex solution is titrated with 0 026 M HCl and the equivalence point. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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]
Thank you for reading the article A 25 0 mL sample of a saturated tex Ca OH 2 tex solution is titrated with 0 026 M HCl and the equivalence point. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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