How does the solubility of `CaC_2O_4` in a 0.1 M solution of `(NH)_4C_2O_4` decrease in comparison with its solubility in water? Assume that the ionisation of `(NH_4)_2C_2O_4` is complete. `K(CaC_2O_4)= 2 xx 10^(-9)`

Answer

To solve the problem of how the solubility of \( \text{CaC}_2\text{O}_4 \) in a 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \) decreases compared to its solubility in water, we will follow these steps:

### Step 1: Determine the solubility of \( \text{CaC}_2\text{O}_4 \) in water

The solubility product constant \( K_{sp} \) for \( \text{CaC}_2\text{O}_4 \) is given as \( 2 \times 10^{-9} \). The dissociation of \( \text{CaC}_2\text{O}_4 \) can be represented as:

\[ \text{CaC}_2\text{O}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{C}_2\text{O}_4^{2-} (aq) \]

If \( s \) is the solubility of \( \text{CaC}_2\text{O}_4 \) in water, then at equilibrium:

\[ [\text{Ca}^{2+}] = s \quad \text{and} \quad [\text{C}_2\text{O}_4^{2-}] = 2s \]

Substituting these into the \( K_{sp} \) expression:

\[ K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]^2 = s(2s)^2 = 4s^3 \]

Setting this equal to the given \( K_{sp} \):

\[ 4s^3 = 2 \times 10^{-9} \]

Solving for \( s \):

\[ s^3 = \frac{2 \times 10^{-9}}{4} = 0.5 \times 10^{-9} = 5 \times 10^{-10} \]

\[ s = \sqrt[3]{5 \times 10^{-10}} \approx 8.0 \times 10^{-4} \, \text{M} \]

### Step 2: Determine the solubility of \( \text{CaC}_2\text{O}_4 \) in a 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \)

In a 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \), the concentration of \( \text{C}_2\text{O}_4^{2-} \) ions is 0.1 M. The equilibrium expression for \( \text{CaC}_2\text{O}_4 \) now becomes:

\[ K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]^2 \]

Let \( s' \) be the solubility of \( \text{CaC}_2\text{O}_4 \) in this solution. Then:

\[ [\text{Ca}^{2+}] = s' \quad \text{and} \quad [\text{C}_2\text{O}_4^{2-}] = 0.1 \]

Substituting into the \( K_{sp} \) expression gives:

\[ 2 \times 10^{-9} = s' (0.1)^2 \]

\[ 2 \times 10^{-9} = s' \times 0.01 \]

Solving for \( s' \):

\[ s' = \frac{2 \times 10^{-9}}{0.01} = 2 \times 10^{-7} \, \text{M} \]

### Step 3: Compare the solubilities

Now we can compare the solubility of \( \text{CaC}_2\text{O}_4 \) in water and in the 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \):

- Solubility in water: \( s \approx 8.0 \times 10^{-4} \, \text{M} \)
- Solubility in 0.1 M \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \): \( s' = 2 \times 10^{-7} \, \text{M} \)

### Step 4: Calculate the decrease in solubility

To find the ratio of the solubility in water to that in the ammonium oxalate solution:

\[ \frac{s}{s'} = \frac{8.0 \times 10^{-4}}{2 \times 10^{-7}} = 4000 \]

This means the solubility of \( \text{CaC}_2\text{O}_4 \) decreases by a factor of 4000 when dissolved in the 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \).

### Conclusion

The solubility of \( \text{CaC}_2\text{O}_4 \) decreases significantly in a 0.1 M solution of \( (\text{NH}_4)_2\text{C}_2\text{O}_4 \) compared to its solubility in pure water due to the common ion effect, where the presence of \( \text{C}_2\text{O}_4^{2-} \) ions suppresses the solubility of \( \text{CaC}_2\text{O}_4 \).

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