# How do you calculate Ksp from solubility?

May 22, 2018

Well, let us seek an example....

#### Explanation:

For $\text{lead(II) chloride}$, this site quotes a solubility of $10.8 \cdot g \cdot {L}^{-} 1$ under standard condition..., and this is a molar solubility of $\frac{\frac{10.8 \cdot g}{278.10 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = 0.03883 \cdot m o l \cdot {L}^{-} 1$ so we address the equilibrium....

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

And in the standard way, we write the $\text{solubility expression...}$, with $S = \text{molar solubility of lead chloride}$

${K}_{\text{sp}} = \left[P {b}^{{2}_{+}}\right] {\left[C {l}^{-}\right]}^{2} = S \times {\left(2 S\right)}^{2} = 4 \times {\left(0.03833\right)}^{3} = 2.28 \times {10}^{-} 4$ ...