Question

How I do the THEORETICAL calculation of solubility?

Answer 1

Well, you would need the solubility product constant `K_(sp)`, and you'd need to identify the compound...

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For some generic ionic compound `M_(nu_(+))X_(nu_(-))`, the dissociation in pure water is given as:

`M_(nu_(+))X_(nu_(-))(s) stackrel(H_2O(l))(rightleftharpoons) nu_(+)M^(z_+)(aq) + nu_(-)X^(z_-)(aq)`

where `nu_(pm)` is the stoichiometric coefficient for the cation `(+)` or anion `(-)`, while `z_pm` is the charge of the cation `(+)` or anion `(-)`.

Its mass action expression would then be:

`K_(sp) = [M^(z_+)]^(nu_(+))[X^(z_-)]^(nu_(-))`

If we define solubility as `s` for the ion that there is `"1 mol"` of, then in general, noting that the coefficients go into the exponents as well as the concentrations:

`K_(sp) = (nu_(+)cdot s)^(nu_(+))(nu_(-)cdot s)^(nu^(-))`

`= nu_(+)^(nu_(+))nu_(-)^(nu_(-))cdot s^(nu_(+) + nu_(-))`

In general, it would mean the solubility is:

`color(blue)(barul|stackrel(" ")(" "s = (K_(sp)/(nu_(+)^(nu_(+))nu_(-)^(nu_(-))))^(1//(nu_(+)+nu_(-)))" ")|)`

If we have, say, `"Ca"_3("PO"_4)_2`, then we write that dissociation as:

`"Ca"_3("PO"_4)_2(s) rightleftharpoons3 "Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq)`

In this case,

`nu_(+) = 3`
`nu_(-) = 2`
`z_(+) = 2`
`z_(-) = -3`

Therefore,

`K_(sp) = ["Ca"^(2+)]^3["PO"_4^(3-)]^2`

`= (3 cdot s)^3 cdot (2 cdot s)^2`

`= 3^3 cdot 2^2 cdot s^(3+2)`

`= 108s^5`

And then its solubility would be given by:

`color(blue)(barul|stackrel(" ")(" "s = (K_(sp)/108)^(1//5)" ")|)`

For `"Ca"_3("PO"_4)_2`, the `K_(sp)` is somewhere around `2.07 xx 10^(-33)`, which indicates that it is extremely insoluble in water at room temperature.

`s = ((2.07 xx 10^(-33))/(108))^(1//5)`

`= 1.14 xx 10^(-7) "M"`

This means that in `"100 mL"` of water at `25^@ "C"`, one could probably dissolve:

`1.14 xx 10^(-8) cancel"mols" xx ("310.172 g Ca"_3("PO"_4)_2)/(cancel"1 mol")`

`= 3.53 xx 10^(-6) "g" ~~ "0.00353 mg"`

However, this `K_(sp)` is apparently not well-known, because I saw various sources list `2.0 xx 10^(-29)`, `2.07 xx 10^(-33)`, and `1.3 xx 10^(-26)`. The second one is the most common.