Let's break down the problem step-by-step to determine the molarity of the HCl solution when it reacts completely with a given amount of Ca(OH)[tex]\(_2\)[/tex].
We are given:
- Volume of Ca(OH)[tex]\(_2\)[/tex] solution: 2.00 liters
- Molarity of Ca(OH)[tex]\(_2\)[/tex] solution: 1.50 M
- Volume of HCl solution: 1.00 liter
The balanced chemical equation is:
[tex]\[ 2 \mathrm{HCl} + \mathrm{Ca(OH)_2} \rightarrow \mathrm{CaCl_2} + 2 \mathrm{H_2O} \][/tex]
From the balanced equation, we see that 1 mole of Ca(OH)[tex]\(_2\)[/tex] reacts with 2 moles of HCl.
### Step 1: Calculate the moles of Ca(OH)[tex]\(_2\)[/tex]
Using the molarity and volume of the Ca(OH)[tex]\(_2\)[/tex] solution, we can calculate the moles of Ca(OH)[tex]\(_2\)[/tex]:
[tex]\[
\text{Moles of Ca(OH)}_2 = (\text{Molarity of Ca(OH)}_2) \times (\text{Volume of Ca(OH)}_2)
\][/tex]
[tex]\[
\text{Moles of Ca(OH)}_2 = 1.50 \, \text{M} \times 2.00 \, \text{L} = 3.00 \, \text{moles}
\][/tex]
### Step 2: Determine the moles of HCl required
From the stoichiometry of the balanced equation, we know:
[tex]\[ 1 \text{ mole of Ca(OH)}_2 \text{ reacts with } 2 \text{ moles of HCl} \][/tex]
Therefore, the total moles of HCl required are:
[tex]\[
\text{Moles of HCl required} = 2 \times (\text{Moles of Ca(OH)}_2)
\][/tex]
[tex]\[
\text{Moles of HCl required} = 2 \times 3.00 \, \text{moles} = 6.00 \, \text{moles}
\][/tex]
### Step 3: Calculate the molarity of the HCl solution
Now, we know the volume of the HCl solution used and the moles of HCl needed. We can calculate the molarity (M) of the HCl solution using:
[tex]\[
\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl in liters}}
\][/tex]
[tex]\[
\text{Molarity of HCl} = \frac{6.00 \, \text{moles}}{1.00 \, \text{L}} = 6.00 \, \text{M}
\][/tex]
### Conclusion
The molarity of the HCl solution is:
[tex]\[
\boxed{6.00 \, \text{M}}
\][/tex]
