1. Oct 19, 2005

### sniffer

what is the product of the reaction:

PbO + NaOH -> ???

or

PbO + NaOH + H2O -> ??

is there any difference?

thanks

2. Oct 19, 2005

### Cesium

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

$$Pb(s) + 2OH^-(aq) \rightarrow Pb(OH)_2(s)$$

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

$$PbO(s) + H2O(l) + OH^-(aq) \rightarrow Pb(OH)_3^-(aq)$$

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

$$PbO2_(s) + 2H_2O(l) + 2OH^-(aq) \rightarrow Pb(OH)_6^{-2}(aq).$$

Last edited: Oct 20, 2005