Question

For the dissolution of lead(II) chloride in `"0.2 M NaCl"`, find the molar solubility at SATP? How do I do this without a `K_(sp)`?

Answer 1

About `4.25 xx 10^(-4)` `"M"` via the small `x` approximation.

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Well, many if not all textbook appendices that list `K_(sp)` list them for `25^@ "C"` and `"1 atm"`, but SATP has `"1 bar"`. We approximate that `K_(sp)` is the same at `"1 bar"` (`"0.987 atm"`) as at `"1 atm"` since molar solubility of solids is negligibly pressure-dependent.

We have

`K_(sp) = 1.7 xx 10^(-5)`

for

`"PbCl"_2(s) rightleftharpoons "Pb"^(2+)(aq) + 2"Cl"^(-)(aq)`.

In a `"0.2 M NaCl"` soln, the conc. of `"Cl"^(-)` is also `"0.2 M"`, which induces a common-ion effect on the `"Cl"^(-)` that dissociates from `"PbCl"_2`, thereby forcing a shift in the equilibrium leftwards from Le Chatelier's Principle.

The ICE table has:

`"PbCl"_2(s) rightleftharpoons "Pb"^(2+)(aq) + 2"Cl"^(-)(aq)`

`"I"" "-" "" "" "" ""0 M"" "" "" ""0.2 M"`
`"C"" "-" "" "" "+s" "" "" "+2s`
`"E"" "-" "" "" "" "s" "" "" "" "(0.2 + 2s) "M"`

And the `K_(sp)` expression therefore is:

`1.7 xx 10^(-5) = s(0.2 + 2s)^2`

`= s(0.04 + 0.8s + 4s^2) = 4s^3 + 0.8s^2 + 0.04s`

where `s` is the molar solubility of `"PbCl"_2` (or `"Pb"^(2+)` I suppose, by a `1:1` mol ratio).

We then get:

`4s^3 + 0.8s^2 + 0.04s - 1.7 xx 10^(-5) = 0`

which would be cumbersome to solve. However, since `K_(sp)` is small (around `10^(-5)` or less), we can say that `2s` is small compared to `0.2`.

`=> 0.2 + 2s ~~ 0.2`

`=> color(blue)(s) ~~ (1.7 xx 10^(-5))/0.2^2 = color(blue)(4.25 xx 10^(-4))` `color(blue)("M")`

(compared to the true answer of `4.21 xx 10^(-4)` `"M"`.)

And `s` is, again, the molar solubility of `"PbCl"_2` when placed in this solution.