# 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?

Nov 12, 2015

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 N a O H \rightarrow P b {\left(O H\right)}_{2} \left(s\right) \downarrow + 2 N a N {O}_{3} \left(a q\right)$

#### 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. $P b {\left(N {O}_{3}\right)}_{2}$: $25.5 \times {10}^{- 3} \cdot L \times 0.150 \cdot m o l \cdot {L}^{- 1}$ $=$ ?? mol.

$N a O H$: $30.0 \times {10}^{- 3} \cdot L \times 0.200 \cdot m o l \cdot {L}^{- 1}$ $=$ ?? 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:

$P {b}^{2 +} + 2 O {H}^{-} \rightarrow P b {\left(O H\right)}_{2} \left(s\right) \downarrow$

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