A white precipitate can be obtained by adding dilute sulphuric acid to :

(a) $C u S O_{4}$ Solution (b) NaCI Solution (c) $B a C l_{2}$ Solution (d) $N a_{2} S O_{4}$ Solution

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Solution

Option (c) is correct.
(a) Reaction of $C u S O_{4} w i t h H_{2} S O_{4}$
$C u S O_{4} + H_{2} S O_{4} ⟶ N o R e a c t i o n$
(b) Reaction of $N a C l w i t h H_{2} S O_{4}$
$2 N a C l + H_{2} S O_{4} ⟶ N a_{2} S O_{4} + 2 H C l$
(c) Reaction of $B a C l_{2} w i t h H_{2} S O_{4}$
$B a C l_{2} + H_{2} S O_{4} ⟶ B a S O_{4} + 2 H C l$
(white ppt)
(d) $R e a c t i o n o f N a_{2} S O_{4} w i t h H_{2} S O_{4}$
$N a_{2} S O_{4} + H_{2} S O_{4} ⟶ N o R e a c t i o n$

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