Question

What is the theoretical yield of solid product when 25.5 ml of 0.150 M lead (II) nitrate is mixed with 30.0 ml of 0.200 M NaOH?

Answer 1

`Pb(NO_3)_2(aq) + 2NaOH rarr Pb(OH)_2(s)darr + 2NaNO_3(aq)`

Explanation:

We have the stoichiometric reaction above. I also happen to know that hydroxides are generally insoluble (hence I can predict that the lead hydroxide will precipitate).

We can find the molar quantities of each reagent. `Pb(NO_3)_2`: `25.5xx10^(-3)*Lxx0.150*mol*L^(-1)` `=` `?? mol`.

`NaOH`: `30.0xx10^(-3)*Lxx0.200*mol*L^(-1)` `=` `?? mol`. Remember that quantitative precipitation of the lead hydroxide requires `2` equivs of hydroxide, as the equation specifies.

I could rewrite the solubility reaction as a net ionic equation by removing the aqueous species:

`Pb^(2+) + 2OH^(-) rarr Pb(OH)_2(s)darr`

The sodium and nitrate ions will remain in solution, and are just along for the ride.