A solution containing 0.01 mole/litre of `CaCl_2` and 0.01 mole/litre of `SrCl_2` is slowly added to a 0.01 N solution of `H_2SO_4` Which substance begins to precipitate earlier?
(a) `SrSO_(4) (b) CaSO_(4). K_(sp) =3.2 xx 10^(-7), K_(sp) (CaSO_(4))=1.3 xx 10^(-4)`.

Answer

To determine which substance begins to precipitate earlier when a solution containing 0.01 mole/litre of CaCl₂ and 0.01 mole/litre of SrCl₂ is slowly added to a 0.01 N solution of H₂SO₄, we need to calculate the ionic products of SrSO₄ and CaSO₄ and compare them with their respective solubility product constants (Ksp).

### Step-by-Step Solution:

1. **Identify the relevant ions and their concentrations**:
- From the given solutions:
- CaCl₂ dissociates into Ca²⁺ and 2Cl⁻, contributing 0.01 M of Ca²⁺.
- SrCl₂ dissociates into Sr²⁺ and 2Cl⁻, contributing 0.01 M of Sr²⁺.
- The H₂SO₄ dissociates into 2H⁺ and SO₄²⁻, contributing 0.01 M of SO₄²⁻.

2. **Calculate the ionic product for SrSO₄**:
- The ionic product (IP) for SrSO₄ is given by:
\[ IP_{SrSO_4} = [Sr^{2+}][SO_4^{2-}] \]
- Substituting the concentrations:
\[ IP_{SrSO_4} = (0.01)(0.01) = 1.0 \times 10^{-4} \]

3. **Compare with Ksp of SrSO₄**:
- The Ksp for SrSO₄ is given as \(3.2 \times 10^{-7}\).
- Since \(IP_{SrSO_4} (1.0 \times 10^{-4}) > Ksp_{SrSO_4} (3.2 \times 10^{-7})\), precipitation of SrSO₄ will begin.

4. **Calculate the ionic product for CaSO₄**:
- The ionic product (IP) for CaSO₄ is given by:
\[ IP_{CaSO_4} = [Ca^{2+}][SO_4^{2-}] \]
- Substituting the concentrations:
\[ IP_{CaSO_4} = (0.01)(0.01) = 1.0 \times 10^{-4} \]

5. **Compare with Ksp of CaSO₄**:
- The Ksp for CaSO₄ is given as \(1.3 \times 10^{-4}\).
- Since \(IP_{CaSO_4} (1.0 \times 10^{-4}) < Ksp_{CaSO_4} (1.3 \times 10^{-4})\), precipitation of CaSO₄ will not begin at this stage.

6. **Conclusion**:
- Since SrSO₄ precipitates when its ionic product exceeds its Ksp before CaSO₄ does, the substance that begins to precipitate earlier is **SrSO₄**.

### Final Answer:
(b) **SrSO₄** begins to precipitate earlier.