lead oxide + base

[sniffer]

what is the product of the reaction:

PbO + NaOH -> ???

or

PbO + NaOH + H2O -> ??

is there any difference?:frown:

thanks

[Cesium]

Ok finally got latex how I wanted. It depends upon concentration of the base. Look up hydrolysis.

Excess Lead:
Pb(s) + 2OH^-(aq) \rightarrow Pb(OH)_2(s)

Excess NaOH:
Pb(OH)_2(s) + H2O(l) \rightarrow Pb(OH)_3^-(aq)

Lead (II) Oxide:
PbO(s) + H2O(l) + OH^-(aq) \rightarrow Pb(OH)_3^-(aq)

So if you use excess NaOH there will be no difference in your products.

Lead (IV) Oxide:
PbO2_(s) + 2H_2O(l) + 2OH^-(aq) \rightarrow Pb(OH)_6^{-2}(aq).