If 20 ml of `2xx10^(-5) BaCl_(2)` solution is mixed with 20 ml of `1xx10^(-5) M Na_(2)SO_(4)` solution, will a ppt. form ? `(K_(sp) "for" BaSO_(4) " is" 1.0xx10^(-10))`

Answer

To determine whether a precipitate will form when 20 mL of \(2 \times 10^{-5} \, \text{M} \, \text{BaCl}_2\) solution is mixed with 20 mL of \(1 \times 10^{-5} \, \text{M} \, \text{Na}_2\text{SO}_4\) solution, we need to follow these steps:

### Step 1: Calculate the moles of BaCl₂ and Na₂SO₄

1. **Calculate moles of BaCl₂:**
\[ \text{Moles of BaCl}_2 = \text{Molarity} \times \text{Volume} = 2 \times 10^{-5} \, \text{mol/L} \times 20 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 4 \times 10^{-7} \, \text{mol} \]

2. **Calculate moles of Na₂SO₄:**
\[ \text{Moles of Na}_2\text{SO}_4 = \text{Molarity} \times \text{Volume} = 1 \times 10^{-5} \, \text{mol/L} \times 20 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 2 \times 10^{-7} \, \text{mol} \]

### Step 2: Determine the total volume of the mixed solution

The total volume after mixing:
\[ \text{Total Volume} = 20 \, \text{mL} + 20 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L} \]

### Step 3: Calculate the concentrations of Ba²⁺ and SO₄²⁻ ions in the mixed solution

1. **Concentration of Ba²⁺:**
\[ \text{Concentration of Ba}^{2+} = \frac{\text{Moles of BaCl}_2}{\text{Total Volume}} = \frac{4 \times 10^{-7} \, \text{mol}}{0.040 \, \text{L}} = 1 \times 10^{-5} \, \text{M} \]

2. **Concentration of SO₄²⁻:**
\[ \text{Concentration of SO}_4^{2-} = \frac{\text{Moles of Na}_2\text{SO}_4}{\text{Total Volume}} = \frac{2 \times 10^{-7} \, \text{mol}}{0.040 \, \text{L}} = 5 \times 10^{-6} \, \text{M} \]

### Step 4: Calculate the ionic product (IP) for BaSO₄

The ionic product (IP) for the formation of BaSO₄ is given by:
\[ \text{IP} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (1 \times 10^{-5})(5 \times 10^{-6}) = 5 \times 10^{-11} \]

### Step 5: Compare the ionic product with the solubility product (Ksp)

Given:
\[ K_{sp} \text{ for BaSO}_4 = 1.0 \times 10^{-10} \]

Now, we compare:
\[ \text{IP} = 5 \times 10^{-11} < K_{sp} = 1.0 \times 10^{-10} \]

### Conclusion

Since the ionic product (IP) is less than the solubility product (Ksp), no precipitate will form.

### Final Answer
**No precipitate will form.**

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