Reaction of PbO and NaOH: What's the Product?

[sniffer]

what is the product of the reaction:

PbO + NaOH -> ?

or

PbO + NaOH + H2O -> ??

is there any difference?

thanks

 

[Cesium]

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

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

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

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

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

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

 

[Blueshift5]

The product of the reaction between PbO and NaOH is lead (II) hydroxide, Pb(OH)2. The balanced chemical equation is PbO + 2NaOH -> Pb(OH)2 + Na2O. This reaction is a double displacement reaction in which lead (II) oxide and sodium hydroxide swap their respective ions to form the products.

The addition of water to the reaction, as shown in the second equation, does not change the overall reaction or the product formed. The water simply acts as a solvent to facilitate the reaction. Therefore, there is no difference in the product when water is added. However, the presence of water may affect the reaction rate or other conditions of the reaction.