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In the quantitative analysis of acidic radicals, when $50 m l$ of $0.1 M K_{2} S O_{4}$ solution is mixed with $50 m l$ of $0.050 M B a C l_{2}$ solutions then, identify the correct statements

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Potassium chloride is precipitated $(k_{s p} = 21.7)$

The $[B a^{+ 2}]$ in the resulting solution will be $0.025 M$

Barium sulphate is precipitate $(k_{s p} = 1.1 \times 10^{- 10})$

The precipitate formed can form pseudo alum with $A l_{2} {(S O_{4})}_{3}$

answer is B, C, D.

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Detailed Solution

$K_{2} S O_{4} + B a C l_{2} \to B a S O_{4} ↓$

$[B a^{+ 2}] \Rightarrow M_{1} V_{1} = M_{2} V_{2} \Rightarrow \frac{50 \times 0.05}{100}$

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