Answer:
[tex]Ksp=3.39x10^{-14}[/tex]
Explanation:
Hello,
In this case, since the dissociation of gold (V) oxalate is:
[tex]Au_2(C_2O_4)_5(s)\rightleftharpoons 2Au^{5+}(aq)+5(C_2O_4)^{2-}(aq)[/tex]
In such a way, the equilibrium expression is:
[tex]Ksp=[Au^{5+}]^2[(C_2O_4)^{2-}]^5[/tex]
Thus, since the molar solubility of the gold (V) oxalate is computed by considering its molar mass (834 g/mol):
[tex][Au_2(C_2O_4)_5]=2.58\frac{g}{L} *\frac{1mol}{834g} =3.09M[/tex]
In such a way, since gold (V) is in a 2:1 molar ratio with the salt and the oxalate in a 5:1 in the chemical reaction, the corresponding concentrations at equilibrium are:
[tex][Au^{5+}]=3.09x10^{-3}\frac{mol}{L} *\frac{2molAu^{5+}}{1mol} =6.19x10^{-3}M[/tex]
[tex][(C_2O_4)^{2-}]=3.09x10^{-3}\frac{mol}{L} *\frac{5mol(C_2O_4)^{2-}}{1mol} =0.0155M[/tex]
Therefore, the solubility product turns out:
[tex]Ksp=(6.19x10^{-3})^2*(0.0155)^5\\\\Ksp=3.39x10^{-14}[/tex]
Regards.
