How I do the THEORETICAL calculation of solubility?

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

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.