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Salt of Weak Acid and Weak Base

Hydrated salt...

Question

Hydrated salt (A) + $B a C l_{2} ⟶$ white ppt (B) insoluble in $H N O_{3}$
(A) (in dil..HCl) + $H_{2} S ⟶$ black ppt soluble in 50% $H N O_{3}$
(A) +aq $N H_{3}$ (excess) $⟶$ deep blue solution (C)
(A) +microscomic salt $△ \to$ green (D) when hot
(A) +Kl solution $⟶$ reddish brown changing to white ppt after hypo is added.
2.495 g of (A) gave 2.33 g of (B). Here (A) is identified as $C u S O_{4} . 5 H_{2} O$ .
If true enter 1, else enter 0.
1

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Solution

The correct option is A 1
(A) + $B a C l_{2}$ solution $⟶$ white ppt (B)
$\Rightarrow S O_{4}^{2 -}$ is present
(A) + $d i l . H C l + H_{2} S ⟶$ black ppt soluble in 50% $H N O_{3}$
$\Rightarrow$ group II (copper subgroup) except $H g^{2 +}$
(A) + aq $N H_{3}$ (excess) $⟶$ deep blue solution (C)
$\Rightarrow C u^{2 +}$ is present
Thus, (A) is $C u S O_{4} . x H_{2})$
$C u S O_{4} . x H_{2} O + B a C l_{2} (159.5 + 18 x) g ⟶ B a S O_{4} 233 g$
$\frac{159.5 + 18 x}{233}$ = $\frac{2.495}{2.33}$
$\Rightarrow$ x= 5
Thus, (A) is $C u S O_{4} . 5 H_{2} O$
Explanation :
$C u S O_{4} + 4 N H_{3} ⟶ [C u (N H_{3})_{4}] S O_{4} d e e p b l u e (C)$
$N a (N H_{4}) H P O_{4} △ \to N a P O_{3} + H_{2} O + N H_{3}$
$N a P O_{3} + C u S O_{4} ⟶ △ \to N a C u P O_{4} g r e e n + S O_{3} ↑$
$2 C u S O_{4} + 4 K l ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$

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