Net ionic reaction of following reaction is :
$P b (N O_{3})_{2} (a q) + (N H_{4})_{2} S (a q) ⟶ P b S (s) + 2 N H_{3} N O_{3} (a q)$

P b^{2 +} (a q) + S^{2 -} (a q) ⟶ P b S (s)

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3 P b^{2 +} (a q) + 2 S^{2 -} (a q) ⟶ 2 P b S (s)

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P b^{2 +} (a q) + 3 S^{2 -} (a q) ⟶ 3 P b S (s)

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none of these

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Solution

The correct option is A $P b^{2 +} (a q) + S^{2 -} (a q) ⟶ P b S (s)$
The given reaction is
$P b (N O_{3})_{2} (a q) + (N H_{4})_{2} S (a q) ⟶ P b S (s) + 2 N H_{4} N O_{3} (a q)$
The complete ionic equation is
$P b^{2 +} (a q) + 2 N O_{3} (a q) + 2 N H_{4} (a q) + S^{2 -} (a q) ⟶ P b S (s) + 2 N H_{4}^{+} (a q) + 2 N O_{3} (a q)$
The net ionic equation is
$P b^{2 +} (a q) + S^{2 -} (a q) ⟶ P b S (s)$

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K_{s p} (P b S O_{4}) = 1.8 \times 10^{- 8}

K_{a} (H S O_{4}^{⊝}) = 1.0 \times 10^{- 2}

P b S O_{4} (s) + H^{\oplus} (a q) ⇌ H S O_{4}^{⊝} (a q) + P b^{2 +} (a q)

S_{2} O {_{6}}^{2 -} (a q) + C r_{2} O {_{7}}^{2 -} (a q) \to S_{4} O {_{6}}^{2 -} (a q) + C r^{3 +} (a q)

K_{s p} (P b S O_{4}) = 1.8 \times 10^{- 8}

K_{a} (H S O_{4}^{⊝}) = 1.0 \times 10^{- 2}

P b S O_{4} (s) + H^{\oplus} (a q) ⇌ H S O_{4}^{⊝} (a q) + P b^{2 +} (a q)

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