Certainly! Let's go through the steps to determine how many milligrams of calcium carbonate (CaCO₃) can dissolve in 200.00 mL of water at 25°C, given the solubility product constant (Ksp) is [tex]\(4.5 \times 10^{-9}\)[/tex].
### Step 1: Writing the Dissociation Equation
Calcium carbonate (CaCO₃) dissociates in water as follows:
[tex]\[ \text{CaCO}_3 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \][/tex]
### Step 2: Solubility Product Constant Expression
The solubility product constant (Ksp) for CaCO₃ is given by:
[tex]\[ K_{\text{sp}} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \][/tex]
Let the solubility of CaCO₃ in water be [tex]\( S \)[/tex] mol/L. At equilibrium, the concentration of both ions will be [tex]\( S \)[/tex]:
[tex]\[ K_{\text{sp}} = S \cdot S = S^2 \][/tex]
### Step 3: Solving for Solubility (S)
[tex]\[ S^2 = 4.5 \times 10^{-9} \][/tex]
[tex]\[ S = \sqrt{4.5 \times 10^{-9}} \][/tex]
[tex]\[ S = 6.71 \times 10^{-5} \text{ mol/L} \][/tex]
### Step 4: Converting Solubility to Grams per Liter
The molar mass (Mr) of CaCO₃ is 100.09 g/mol. Therefore, the solubility in g/L is:
[tex]\[ \text{Solubility (g/L)} = S \times \text{Mr} \][/tex]
[tex]\[ \text{Solubility (g/L)} = 6.71 \times 10^{-5} \text{ mol/L} \times 100.09 \text{ g/mol} \][/tex]
[tex]\[ \text{Solubility (g/L)} = 0.006714 \text{ g/L} \][/tex]
### Step 5: Converting Solubility to the Given Volume of Water
We need the total mass of CaCO₃ in 200.00 mL (0.200 L) of water:
[tex]\[ \text{Total mass (g)} = \text{Solubility (g/L)} \times \text{Volume (L)} \][/tex]
[tex]\[ \text{Total mass (g)} = 0.006714 \text{ g/L} \times 0.200 \text{ L} \][/tex]
[tex]\[ \text{Total mass (g)} = 0.0013428 \text{ g} \][/tex]
### Step 6: Converting Grams to Milligrams
Since 1 gram equals 1000 milligrams:
[tex]\[ \text{Total mass (mg)} = 0.0013428 \text{ g} \times 1000 \][/tex]
[tex]\[ \text{Total mass (mg)} = 1.3428 \text{ mg} \][/tex]
Therefore, the amount of CaCO₃ that can dissolve in 200.00 mL of water at 25°C is approximately 1.3428 milligrams.
