How many grams of calcium oxalate should be dissolved in water to make one litre of saturated solution? `K_sp` of `CaC_2O_4` is `2.5 x 10^-9` and its molecular weight is 128 u

Options

A. `6.4*10^-3 g`
B. `8.0*10^-3 g`
C. `1.28*10^-3 g`
D. `6.4*3.2^-3 g`

Answer

To determine how many grams of calcium oxalate (CaC₂O₄) should be dissolved in water to make one liter of a saturated solution, we can follow these steps:

### Step 1: Write the dissociation equation
Calcium oxalate dissociates in water as follows:
\[ \text{CaC}_2\text{O}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{C}_2\text{O}_4^{2-} (aq) \]

### Step 2: Define solubility (S)
Let the solubility of calcium oxalate in moles per liter be \( S \). Therefore, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) ions will be \( S \).
- The concentration of \( \text{C}_2\text{O}_4^{2-} \) ions will also be \( S \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant (\( K_{sp} \)) for calcium oxalate is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}] \]
Substituting the concentrations:
\[ K_{sp} = S \cdot S = S^2 \]

### Step 4: Substitute the value of \( K_{sp} \)
Given that \( K_{sp} = 2.5 \times 10^{-9} \):
\[ S^2 = 2.5 \times 10^{-9} \]

### Step 5: Solve for S
To find \( S \), take the square root of both sides:
\[ S = \sqrt{2.5 \times 10^{-9}} \]
\[ S = 5 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Convert moles to grams
Now, we need to convert the solubility from moles to grams using the molecular weight of calcium oxalate, which is 128 g/mol:
\[ \text{Mass} = S \times \text{Molecular Weight} \]
\[ \text{Mass} = 5 \times 10^{-5} \, \text{mol/L} \times 128 \, \text{g/mol} \]
\[ \text{Mass} = 6.4 \times 10^{-3} \, \text{g} \]
\[ \text{Mass} = 0.0064 \, \text{g} \]

### Final Answer
To prepare one liter of a saturated solution of calcium oxalate, you should dissolve approximately **0.0064 grams** of calcium oxalate.

---