How many grams of `MgC_(2)O_(4)` (formula mass =112) will dissolve in 1.5 L of water?
`(K_(sp)=8.1xx10^(-5))`

Options

A. `1.0`
B. `1.29`
C. `1.512`
D. `4.65`

Answer

To solve the problem of how many grams of \( \text{MgC}_2\text{O}_4 \) will dissolve in 1.5 L of water given the solubility product constant \( K_{sp} = 8.1 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the Dissolution Reaction
The dissolution of \( \text{MgC}_2\text{O}_4 \) in water can be represented as:
\[ \text{MgC}_2\text{O}_4 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{C}_2\text{O}_4^{2-} (aq) \]

### Step 2: Define Solubility
Let the solubility of \( \text{MgC}_2\text{O}_4 \) be \( S \) (in mol/L). From the dissolution reaction, we can see that:
- The concentration of \( \text{Mg}^{2+} \) ions will be \( S \).
- The concentration of \( \text{C}_2\text{O}_4^{2-} \) ions will be \( 2S \).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{C}_2\text{O}_4^{2-}]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the Value of \( K_{sp} \)
Now, we can set up the equation:
\[ 4S^3 = 8.1 \times 10^{-5} \]
To find \( S \), we rearrange the equation:
\[ S^3 = \frac{8.1 \times 10^{-5}}{4} = 2.025 \times 10^{-5} \]

### Step 5: Calculate \( S \)
Taking the cube root:
\[ S = \sqrt[3]{2.025 \times 10^{-5}} \approx 2.68 \times 10^{-2} \, \text{mol/L} \]

### Step 6: Calculate Moles in 1.5 L of Water
Now, we can find the number of moles of \( \text{MgC}_2\text{O}_4 \) that can dissolve in 1.5 L of water:
\[ \text{Moles} = S \times \text{Volume} = (2.68 \times 10^{-2} \, \text{mol/L}) \times (1.5 \, \text{L}) \approx 4.02 \times 10^{-2} \, \text{mol} \]

### Step 7: Calculate Mass of \( \text{MgC}_2\text{O}_4 \)
Finally, we can calculate the mass using the molar mass:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = (4.02 \times 10^{-2} \, \text{mol}) \times (112 \, \text{g/mol}) \approx 4.49 \, \text{g} \]

### Final Answer
Approximately **4.49 grams** of \( \text{MgC}_2\text{O}_4 \) will dissolve in 1.5 L of water.

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