Given that solubility product of `BaSO_(4)` is `1 xx 10^(-10)` will be precipiate from when
Equal volumes of `2 xx 10^(-3)M BaCl_(2)` solution and `2 xx 10^(-4)M Na_(2)SO_(4)` solution, are mixed?

Explanation

To determine whether a precipitate of BaSO₄ will form when equal volumes of 2 x 10⁻³ M BaCl₂ and 2 x 10⁻⁴ M Na₂SO₄ solutions are mixed, we can follow these steps:

### Step 1: Determine the concentrations of Ba²⁺ and SO₄²⁻ ions after mixing.

1. **Calculate the concentration of Ba²⁺ ions from BaCl₂:**
- BaCl₂ dissociates completely in solution:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
- The initial concentration of BaCl₂ is 2 x 10⁻³ M. Since equal volumes are mixed, the concentration of Ba²⁺ after mixing will be:
\[ [\text{Ba}^{2+}] = \frac{2 \times 10^{-3}}{2} = 1 \times 10^{-3} \, \text{M} \]

2. **Calculate the concentration of SO₄²⁻ ions from Na₂SO₄:**
- Na₂SO₄ dissociates completely in solution:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \]
- The initial concentration of Na₂SO₄ is 2 x 10⁻⁴ M. Since equal volumes are mixed, the concentration of SO₄²⁻ after mixing will be:
\[ [\text{SO}_4^{2-}] = \frac{2 \times 10^{-4}}{2} = 1 \times 10^{-4} \, \text{M} \]

### Step 2: Calculate the ionic product (IP) of BaSO₄.

The ionic product (IP) for the precipitation of BaSO₄ can be calculated using the concentrations of Ba²⁺ and SO₄²⁻ ions:
\[ \text{IP} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (1 \times 10^{-3})(1 \times 10^{-4}) = 1 \times 10^{-7} \]

### Step 3: Compare the ionic product with the solubility product (Ksp).

The solubility product (Ksp) of BaSO₄ is given as:
\[ K_{sp} = 1 \times 10^{-10} \]

Now, we compare the ionic product with the solubility product:
- Since \( \text{IP} = 1 \times 10^{-7} \) is greater than \( K_{sp} = 1 \times 10^{-10} \), a precipitate will form.

### Conclusion:

Thus, when equal volumes of 2 x 10⁻³ M BaCl₂ and 2 x 10⁻⁴ M Na₂SO₄ solutions are mixed, a precipitate of BaSO₄ will form.

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