Question

The solubility of Pb(OH)2 in water is 6.7 *10^-6 M.What is its solubility in a buffer solution of pH=8 ?

Answer 1

For the purposes of this question, the buffer solution may just imply that this pH is fairly constant. For a more precise answer if it doesn't, I would need the buffer constituents.

We're considering this equilibrium,

`Pb(OH)_2(s) rightleftharpoons Pb^(2+)(aq) + 2OH^(-)(aq)`

where `K_("sp") = [Pb^(2+)][OH^-]^2`, and

`K_("sp") approx (6.7*10^-6)(2*6.7*10^-6)^2 approx 1.20*10^-15`

If `"pH" = 8`, then `[OH^(-)] approx 10^-6`.

Focusing again on the equilibrium,

`Pb(OH)_2(s) rightleftharpoons Pb^(2+)(aq) + 2OH^(-)(aq)`

then,

`1.20*10^-15 = x(10^-6+2x)^2`,
`therefore x approx 6.4*10^-6`

assuming that no interaction occurs with the buffer solution. Refer to the answer above if you need something more precise.

Answer 2

I got about `1.20 xx 10^(-3) "M"` immediately after the buffer consumes `"OH"^(-)` from `"Pb"("OH")_2(s)`, assuming `Delta"pH" ~~ 0` (meaning the equilibrium shift leftwards due to the buffer action was not pursued quantitatively).

Past this point, the equilibrium would shift to decrease the solubility, but without the actual buffer components, this remains a qualitative remark.

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WITHOUT BUFFER

At a `"pH"` of `8`, the `"pOH"`, assuming the solution is at `25^@ "C"`, is `14 - "pH" = 6`, so the `["OH"^(-)]` concentration starts at:

`["OH"^(-)]_i = 10^(-"pOH") = 10^(-6) "M"`.

In regular water, the solubility `s` allows us to get `K_(sp)`:

`K_(sp) = ["Pb"^(2+)]["OH"^(-)]^2 = s(2s)^2`

`= 4s^3`

`= 4 xx (6.7 xx 10^(-6) "M")^3`

`= 1.20 xx 10^(-15)`

WITH BUFFER (BEFORE LE CHATELIER SHIFT)

The new solubility after we add `"Pb"("OH")_2(s)` into a new buffer solution is then defined as `s'` in the following equilibrium:

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

`"I"" "" "" "" "" "" "" ""0 M"" "" "" "10^(-6) "M"`
`"C"" "" "" "" "" "" "" "+s'" "" "" "+2s'`
`"E"" "" "" "" "" "" "" "s'" M"" "(10^(-6) + 2s')"M"`

This `K_(sp)` stays the same regardless of what is in solution.

`K_(sp) = s'(10^(-6) + 2s')^2`

This `K_(sp)` is quite small, but it isn't necessarily the case that `2s'` is small compared to `10^(-6) "M"`. However, since the solution is buffered at `"pH" = 8`, it was made to resist changes in `"pH"`.

By adding more `"OH"^(-)` from the dissociation of `"Pb"("OH")_2(s)`, it reacts with weak acid in the buffer to form weak conjugate base.

`"HA"(aq) + "OH"^(-)(aq) -> "A"^(-)(aq) + "H"_2"O"(l)`

That consumes the `2s'` `"M"` of `"OH"^(-)` to slightly increase the `"pH"`, whatever it becomes.

Without exact concentrations of these components of the buffer, we don't know by how much, but it is small because `"Pb"("OH")_2(s)` is far from a strong base.

So we assume the `"pH"` is still close to `8`, and that `2s'` approaches zero after reacting with the buffer.

This gives:

`1.20 xx 10^(-15) ~~ s'(10^(-6))^2`

`=> color(blue)(s' ~~ 1.20 xx 10^(-3) "M")`

LE CHATELIER SHIFT (QUALITATIVE)

Afterwards, these become the current concentrations of `"Pb"^(2+)` and `"OH"^(-)` in this non-equilibrium state:

`["Pb"^(2+)]_(i2) = 1.20 xx 10^(-3) "M"`
`["OH"^(-)]_(i2) ~~ 10^(-6) "M"`

With this, `Q_(sp) = ["Pb"^(2+)]_(i2)["OH"^(-)]_(i2)^2 = 1.20 xx 10^(-9)`, so `Q_(sp) > K_(sp)` and the equilibrium should proceed leftwards, if we assume the buffer negligibly interferes at this point.