College

50.0 mL of an HCl solution with a pH of 3.5 neutralizes 200.0 mL of a [tex]Ca(OH)_2[/tex] solution. What is the molarity of the [tex]Ca(OH)_2[/tex] solution?

1. [tex]7.91 \times 10^{-5} \, M[/tex]
2. [tex]14.3 \, M[/tex]
3. [tex]3.95 \times 10^{-5} \, M[/tex]
4. [tex]4.55 \times 10^{-6} \, M[/tex]
5. [tex]1.58 \times 10^{-4} \, M[/tex]
6. [tex]1.69 \, M[/tex]
7. [tex]0.0342 \, M[/tex]
8. [tex]2.11 \times 10^{-5} \, M[/tex]
9. [tex]1.78 \times 10^{-4} \, M[/tex]

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]