# What is the solubility (in M) of PbCl2 in a 0.15 M solution of HCl? The Ksp of PbCl2 is 1.6 × 10-5?

Nov 9, 2015

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$P b C {l}_{2} r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 C {l}^{-}$; ${K}_{s p} = 1.6 \times {10}^{- 5}$

Solubility of $P b C {l}_{2}$ $=$ $7.11 \times {10}^{- 4}$ $m o l \cdot {L}^{- 1}$.

#### Explanation:

Now here ${K}_{s p} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2}$. In pure water, we would write ${K}_{s p} = \left(S\right) {\left(2 S\right)}^{2} = 4 {S}^{3}$, where $S$ is the solubility of lead chloride.

However, here the concentration of chloride anion has been (artificially) increased by the presence of hydrochloric acid, which gives stoichiometric quantities of $C {l}^{-}$. So $\left[C {l}^{-}\right]$ $=$ (0.15*mol*L^(-1) + the twice the solubility of lead chloride).

Our revised ${K}_{s p}$ expression is ${K}_{s p} = \left\{S\right\} {\left\{0.15 + 2 S\right\}}^{2}$. Now we can (reasonably) make the approximation, $\left\{0.15 + 2 S\right\} \cong 0.15$; we have to justify this approximation later.

So now, ${K}_{s p} \cong \left[S\right] {\left[0.15\right]}^{2}$, and $S = {K}_{s p} / {\left(0.15\right)}^{2}$ $=$ $\frac{1.6 \times {10}^{- 5}}{0.15} ^ 2$ $=$ $7.11 \times {10}^{- 4}$ $m o l \cdot {L}^{- 1}$.

This value is indeed small compared to 0.15. We could make a 2nd approximation and plug this $S$ value back in the first equation but I am not going to bother. Would the solubility of lead chloride be greater in pure water? Why or why not?