To compute the concentrations of \([Ca^{2+}]\) and \([OH^{-}]\) in a solution prepared by dissolving \(0.60 \, g\) of \(Ca(OH)_{2}\) in \(1500 \, mL\) of water, we will follow these steps:
### Step 1: Calculate the molar mass of \(Ca(OH)_{2}\)
The molar mass can be calculated using the atomic masses provided:
- Calcium (Ca) = 40 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
The formula for calcium hydroxide is \(Ca(OH)_{2}\). Thus, the molar mass is calculated as follows:
\[
\text{Molar mass of } Ca(OH)_{2} = 40 + 2(16) + 2(1) = 40 + 32 + 2 = 74 \, g/mol
\]
### Step 2: Calculate the number of moles of \(Ca(OH)_{2}\)
Using the mass of \(Ca(OH)_{2}\) and its molar mass, we can find the number of moles:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.60 \, g}{74 \, g/mol} \approx 0.0081 \, mol
\]
### Step 3: Convert the volume of the solution from mL to L
The volume of the solution is given as \(1500 \, mL\). We need to convert this to liters:
\[
\text{Volume in liters} = \frac{1500 \, mL}{1000} = 1.5 \, L
\]
### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.0081 \, mol}{1.5 \, L} \approx 0.0054 \, M
\]
### Step 5: Determine the concentrations of \([Ca^{2+}]\) and \([OH^{-}]\)
When \(Ca(OH)_{2}\) dissolves in water, it dissociates completely according to the following equation:
\[
Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}
\]
From the dissociation, we can see that:
- For every 1 mole of \(Ca(OH)_{2}\), 1 mole of \(Ca^{2+}\) is produced.
- For every 1 mole of \(Ca(OH)_{2}\), 2 moles of \(OH^{-}\) are produced.
Thus, the concentrations will be:
- \([Ca^{2+}] = 0.0054 \, M\)
- \([OH^{-}] = 2 \times 0.0054 \, M = 0.0108 \, M\)
### Final Answers:
- \([Ca^{2+}] = 5.4 \times 10^{-3} \, M\)
- \([OH^{-}] = 1.08 \times 10^{-2} \, M\)
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