# How do we find the K_"sp" value for "lead nitrate"?

Apr 11, 2017

You need to quote some units.........

#### Explanation:

We interrogate the equilibrium:

$P b {\left(N {O}_{3}\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 N {O}_{3}^{-}$

And ${K}_{\text{sp}} = \left[P {b}^{2 +}\right] {\left[N {O}_{3}^{-}\right]}^{2}$

If we call the solubility of $\text{lead nitrate}$ $=$ $S$, then.......

${K}_{\text{sp}} = \left[P {b}^{2 +}\right] {\left[N {O}_{3}^{-}\right]}^{2} = \left(S\right) {\left(2 S\right)}^{2} = 4 {S}^{3}$.

But you have given us NO VALUE for solubility. What are the units, $\text{grams per mL,}$ $m o l \cdot {L}^{-} 1$, $\text{pounds per pint}$? Of course, I could look up this value, but why should I bother if you can't be bothered.