a. Predict the products formed when ` Pb_(3) O_(4)` reacts with concentrated hydrochloric acid .
b. In which of the acid lead (II)oxide will dissolve : `H_(2) SO_(4)` ro `HNO_3`. Give reason .
c. Give the reaction between (i) `HCl` and `PbO_2 (ii) SO_4` and `PbO_2` .

Explanation

`Pb_(3)O_(4)` consists of `2PO.PbO_(2)`. Oxidation state of `Pb` in
`PbO_(2)` is `+4`, hence, `PbO_(2)` will behave as an oxidsing agent and will oxidise `Hclrarr Cl_(2)`.
`Pb_(3)O_(4)+8HClrarr 3PPbCl_(2) +Cl_(2)+4H_(2)O`
b. Lead (II) oxide reacts with `H_(2)SO_(4)` to faorm lead (II) sulphate which is insoluble, whereas with `NHio_(3)`, it forms lead (II) nitrate which is soluble in water.
`PbO+H_(2)SO_(4) rarr underset(("white ppt."))underset("Lead nitrate)(PbSO_(4)darr)+H_(2)O`
PbO+2HNO_(3)rarrunderset(("soluble"))underset("Lead nitrate")(Pb(NO_(3))_(2)+H_(2)H`
c. i. overset((+4))(PbO_(2))+overset((-1))(HCl)rarroverset((+2))(PbCl_(2))+H_(2)Ooverset((0))(+)Cl_(2)`
In `PbO_(2), Pb` is in `+4` oxidation state, which being less stable than `+2` oxidation state, gets reduced to `+2` oxidation state. `PbO_(2)` behaves as a strong oxidasing agent and oxidises `HCl rarr Cl_(2)`
ii. `overset((+4))(PbO_(2))+SO_(2) rarr overset((+4))(PbSO_(4))`
In `PbO_(2), Pb` is in `+4` oxidation state, which is less stable `+2` oxidation state. Thus, `PbO_(2)` behave as a strong oxidising agent and oxidises `SO_(2)` to `SO_(4)^(2-)`.