Equal volumes of 0.02 N solutions of `CaCl_2 and Na_2SO_4` are mixed. Will there be a formation of `CaSO_4`, precipitate? `K_(sp)(CaSO_4) =1.3 xx 10^(-4)`.

Answer

To determine whether a precipitate of calcium sulfate (CaSO₄) will form when equal volumes of 0.02 N solutions of CaCl₂ and Na₂SO₄ are mixed, we can follow these steps:

### Step 1: Calculate the concentrations after mixing
When equal volumes of two solutions are mixed, the concentrations of the ions will be halved.

Given:
- Initial concentration of CaCl₂ = 0.02 N
- Initial concentration of Na₂SO₄ = 0.02 N

After mixing equal volumes:
- New concentration of Ca²⁺ ions from CaCl₂ = 0.02 N / 2 = 0.01 N
- New concentration of SO₄²⁻ ions from Na₂SO₄ = 0.02 N / 2 = 0.01 N

### Step 2: Calculate the ionic product (IP)
The ionic product (IP) for the formation of CaSO₄ can be calculated using the concentrations of the ions:

\[ \text{IP} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \]
\[ \text{IP} = (0.01)(0.01) = 0.0001 = 1.0 \times 10^{-4} \]

### Step 3: Compare the ionic product with the solubility product (Ksp)
The solubility product (Ksp) of CaSO₄ is given as:

\[ K_{sp}(CaSO_4) = 1.3 \times 10^{-4} \]

Now we compare the ionic product with Ksp:

- IP = \( 1.0 \times 10^{-4} \)
- Ksp = \( 1.3 \times 10^{-4} \)

### Step 4: Conclusion
Since the ionic product (1.0 × 10⁻⁴) is less than the Ksp (1.3 × 10⁻⁴), no precipitate of CaSO₄ will form.

### Final Answer
No, a precipitate of CaSO₄ will not form.

---