$K_{2} C r O_{4}$ and $N a_{2} C r O_{4}$ can be used to identify $B a^{2 +}$ and $P b^{2 +}$ both.
If true enter 1, if false enter 0.

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Solution

Ba(II) and Pb(II) ions react with potassium chromate to give yellow precipitate of metal chromate.
$P b^{2 +} + K_{2} C r O_{4} \to P b C r O_{4} ↓ y e l l o w p r e c i p i t a t e + 2 K^{+}$
$B a^{2 +} + K_{2} C r O_{4} \to B a C r O_{4} ↓ y e l l o w p r e c i p i t a t e + 2 K^{+}$
Similarly, Ba(II) and Pb(II) ions react with sodium sulphate to form white precipitate of metal sulphate.
$P b^{2 +} + N a_{2} S O_{4} \to P b S O_{4} ↓ w h i t e p r e c i p i t a t e + 2 N a^{+}$
$B a^{2 +} + N a_{2} S O_{4} \to B a S O_{4} ↓ w h i t e p r e c i p i t a t e + 2 N a^{+}$

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