Question

The solubility of gold (V) oxalate, Au2(C2O4)5 is 2.58 g/L. Calculate Ksp from this information.

Answer 1

Answer:

$Ksp=3.39x10^{-14}$

Explanation:

Hello,

In this case, since the dissociation of gold (V) oxalate is:

$Au_2(C_2O_4)_5(s)\rightleftharpoons 2Au^{5+}(aq)+5(C_2O_4)^{2-}(aq)$

In such a way, the equilibrium expression is:

$Ksp=[Au^{5+}]^2[(C_2O_4)^{2-}]^5$

Thus, since the molar solubility of the gold (V) oxalate is computed by considering its molar mass (834 g/mol):

$[Au_2(C_2O_4)_5]=2.58\frac{g}{L} *\frac{1mol}{834g} =3.09M$

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:

$[Au^{5+}]=3.09x10^{-3}\frac{mol}{L} *\frac{2molAu^{5+}}{1mol} =6.19x10^{-3}M$

$[(C_2O_4)^{2-}]=3.09x10^{-3}\frac{mol}{L} *\frac{5mol(C_2O_4)^{2-}}{1mol} =0.0155M$

Therefore, the solubility product turns out:

$Ksp=(6.19x10^{-3})^2*(0.0155)^5\\\\Ksp=3.39x10^{-14}$

Regards.