50 ml of 0.04 M calcium nitrate solution is added to 150 ml of 0.08 M ammonium sulphate. Will a precipitate of calcium sulphate form or not ? `K_(sp)=0.4xx10^(-5)`

Explanation

To determine whether a precipitate of calcium sulfate (CaSO₄) will form when 50 mL of 0.04 M calcium nitrate solution is mixed with 150 mL of 0.08 M ammonium sulfate, we will follow these steps:

### Step 1: Calculate the total volume of the mixed solution.
The total volume after mixing is:
\[ \text{Total Volume} = 50 \, \text{mL} + 150 \, \text{mL} = 200 \, \text{mL} \]

### Step 2: Calculate the concentration of calcium ions (Ca²⁺) in the mixed solution.
First, we need to find the number of moles of Ca²⁺ from the calcium nitrate solution:
\[ \text{Moles of Ca(NO₃)₂} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.050 \, \text{L} \times 0.04 \, \text{mol/L} = 0.002 \, \text{mol} \]
Since calcium nitrate dissociates into one Ca²⁺ ion per formula unit, the moles of Ca²⁺ is also 0.002 mol.

Now, we calculate the concentration of Ca²⁺ in the total volume:
\[ \text{Concentration of Ca²⁺} = \frac{\text{Moles of Ca²⁺}}{\text{Total Volume (L)}} = \frac{0.002 \, \text{mol}}{0.200 \, \text{L}} = 0.01 \, \text{M} \]

### Step 3: Calculate the concentration of sulfate ions (SO₄²⁻) in the mixed solution.
Next, we find the number of moles of sulfate ions from ammonium sulfate:
\[ \text{Moles of (NH₄)₂SO₄} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.150 \, \text{L} \times 0.08 \, \text{mol/L} = 0.012 \, \text{mol} \]
Ammonium sulfate dissociates into two ammonium ions and one sulfate ion, so the moles of SO₄²⁻ is:
\[ \text{Moles of SO₄²⁻} = 0.012 \, \text{mol} \]

Now, we calculate the concentration of SO₄²⁻ in the total volume:
\[ \text{Concentration of SO₄²⁻} = \frac{\text{Moles of SO₄²⁻}}{\text{Total Volume (L)}} = \frac{0.012 \, \text{mol}}{0.200 \, \text{L}} = 0.06 \, \text{M} \]

### Step 4: Calculate the ionic product (IP) for calcium sulfate.
The ionic product (IP) is calculated using the concentrations of Ca²⁺ and SO₄²⁻:
\[ \text{IP} = [\text{Ca}^{2+}][\text{SO₄}^{2-}] = (0.01)(0.06) = 0.0006 = 6 \times 10^{-4} \]

### Step 5: Compare the ionic product (IP) with the solubility product (Ksp).
The given Ksp for calcium sulfate is:
\[ K_{sp} = 0.4 \times 10^{-5} = 4 \times 10^{-6} \]

Now we compare the ionic product with Ksp:
\[ \text{IP} = 6 \times 10^{-4} \quad \text{and} \quad K_{sp} = 4 \times 10^{-6} \]

Since \( \text{IP} > K_{sp} \), a precipitate of calcium sulfate will form.

### Conclusion
Yes, a precipitate of calcium sulfate will form.

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