150 mL of 0.0008 M ammonium sulphate is mixed with 50 mL of 0.04 M calcium nitrate. The ionic product of `CaSO_(4)` will be : `(K_(sp)=2.4xx10^(-5) for CaSO_(4))`

Options

A. `ltK_(sp)`
B. `gtK_(sp)`
C. `approxK_(sp)`
D. None of these

Answer

To solve the problem, we need to calculate the ionic product of calcium sulfate (CaSO₄) after mixing two solutions: ammonium sulfate and calcium nitrate. Here are the steps to arrive at the solution:

### Step 1: Calculate the number of moles of calcium ions (Ca²⁺) in the calcium nitrate solution.

Given:
- Volume of calcium nitrate solution (V₁) = 50 mL = 0.050 L
- Molarity of calcium nitrate solution (M₁) = 0.04 M

Using the formula for moles:
\[ \text{Moles of Ca}^{2+} = M₁ \times V₁ \]
\[ \text{Moles of Ca}^{2+} = 0.04 \, \text{mol/L} \times 0.050 \, \text{L} = 0.002 \, \text{mol} \]

### Step 2: Calculate the final concentration of calcium ions (Ca²⁺) after mixing.

Total volume after mixing:
\[ V_{\text{total}} = 150 \, \text{mL} + 50 \, \text{mL} = 200 \, \text{mL} = 0.200 \, \text{L} \]

Using the number of moles calculated in Step 1:
\[ \text{Concentration of Ca}^{2+} = \frac{\text{Moles of Ca}^{2+}}{V_{\text{total}}} \]
\[ \text{Concentration of Ca}^{2+} = \frac{0.002 \, \text{mol}}{0.200 \, \text{L}} = 0.01 \, \text{M} \]

### Step 3: Calculate the number of moles of sulfate ions (SO₄²⁻) in the ammonium sulfate solution.

Given:
- Volume of ammonium sulfate solution (V₂) = 150 mL = 0.150 L
- Molarity of ammonium sulfate solution (M₂) = 0.0008 M

Using the formula for moles:
\[ \text{Moles of SO₄}^{2-} = M₂ \times V₂ \]
\[ \text{Moles of SO₄}^{2-} = 0.0008 \, \text{mol/L} \times 0.150 \, \text{L} = 0.00012 \, \text{mol} \]

### Step 4: Calculate the final concentration of sulfate ions (SO₄²⁻) after mixing.

Using the number of moles calculated in Step 3:
\[ \text{Concentration of SO₄}^{2-} = \frac{\text{Moles of SO₄}^{2-}}{V_{\text{total}}} \]
\[ \text{Concentration of SO₄}^{2-} = \frac{0.00012 \, \text{mol}}{0.200 \, \text{L}} = 6 \times 10^{-4} \, \text{M} \]

### Step 5: Calculate the ionic product (Q) of calcium sulfate (CaSO₄).

The ionic product is given by:
\[ Q = [\text{Ca}^{2+}][\text{SO₄}^{2-}] \]
Substituting the concentrations calculated:
\[ Q = (0.01)(6 \times 10^{-4}) \]
\[ Q = 6 \times 10^{-6} \]

### Step 6: Compare the ionic product (Q) with the solubility product (Ksp).

Given:
- \( K_{sp} \) for CaSO₄ = \( 2.4 \times 10^{-5} \)

Since \( Q = 6 \times 10^{-6} \) and \( K_{sp} = 2.4 \times 10^{-5} \):
\[ Q < K_{sp} \]

### Conclusion:
The ionic product of calcium sulfate is less than the solubility product, indicating that no precipitation will occur.

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