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A 25.0 mL sample of a saturated [tex]$Ca(OH)_2$[/tex] solution is titrated with 0.028 M HCl, and the equivalence point is reached after 38.1 mL of titrant are dispensed. Based on this data, what is the concentration [tex]$(M)$[/tex] of [tex]$Ca(OH)_2$[/tex]?

Type your answer:
[tex]\square[/tex]

Answer :

We start with the titration of a 25.0 mL sample of saturated calcium hydroxide, [tex]$\text{Ca(OH)}_2$[/tex], with 0.028 M hydrochloric acid, [tex]$\text{HCl}$[/tex]. The reaction involved is:

[tex]$$
\text{Ca(OH)}_2 + 2\,\text{HCl} \to \text{CaCl}_2 + 2\,\text{H}_2\text{O}
$$[/tex]

Step 1. Calculate the moles of [tex]$\text{HCl}$[/tex] used.

Convert the volume of [tex]$\text{HCl}$[/tex] from milliliters to liters:

[tex]$$
V_{\text{HCl}} = 38.1\,\text{mL} = 0.0381\,\text{L}
$$[/tex]

Now, calculate the moles of [tex]$\text{HCl}$[/tex]:

[tex]$$
n_{\text{HCl}} = C_{\text{HCl}} \times V_{\text{HCl}} = 0.028\,\text{M} \times 0.0381\,\text{L} = 0.0010668\,\text{mol}
$$[/tex]

Step 2. Determine the moles of [tex]$\text{Ca(OH)}_2$[/tex].

Based on the balanced chemical reaction, 1 mole of [tex]$\text{Ca(OH)}_2$[/tex] reacts with 2 moles of [tex]$\text{HCl}$[/tex]. Therefore, the moles of [tex]$\text{Ca(OH)}_2$[/tex] are half of the moles of [tex]$\text{HCl}$[/tex]:

[tex]$$
n_{\text{Ca(OH)}_2} = \frac{n_{\text{HCl}}}{2} = \frac{0.0010668\,\text{mol}}{2} = 0.0005334\,\text{mol}
$$[/tex]

Step 3. Calculate the concentration of [tex]$\text{Ca(OH)}_2$[/tex].

First, convert the sample volume from mL to L:

[tex]$$
V_{\text{sample}} = 25.0\,\text{mL} = 0.025\,\text{L}
$$[/tex]

Now, calculate the concentration using the formula:

[tex]$$
[\text{Ca(OH)}_2] = \frac{n_{\text{Ca(OH)}_2}}{V_{\text{sample}}} = \frac{0.0005334\,\text{mol}}{0.025\,\text{L}} = 0.021336\,\text{M}
$$[/tex]

Thus, the concentration of the saturated [tex]$\text{Ca(OH)}_2$[/tex] solution is:

[tex]$$
\boxed{0.021336\,\text{M}}
$$[/tex]

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