# What is the difference between carbon dioxide, and silicon dioxide? How do we formulate the solubility product for the salt A_2B_3

Apr 13, 2016

1. Carbon dioxide is a molecular species; silicon dioxide is non-molecular.
2. ${K}_{s p} = {\left[A\right]}^{2} {\left[B\right]}^{3}$

#### Explanation:

As you know, carbon dioxide and silicon dioxide are isoelectronic. They are not isostructural. The shortness of the (first-row) $C - O$ bonds allows effective overlap between the p -orbitals of $C$ and $O$, in addition to the strong $\sigma$ bond formed. The result is a $C = O$ double bond. On the other hand, the corresponding p -orbitals on second row $S i$ are too diffuse to allow effective $\pi$ bonding between $S i$ and $O$, with the result that $O$ bridges to another silicon to form an infinite array of $S i - O - S i - O$ bridges that have no molecular boundary.

As to your ${A}_{2} {B}_{3}$ salt, do you mean a salt of the form such as calcium phosphate, i.e. $C {a}_{3} {\left(P {O}_{4}\right)}_{2}$? Such a salt would be particularly insoluble (as are most phosphates).

We would write the equation for its dissolution in water for as:

$C {a}_{3} {\left(P {O}_{4}\right)}_{2} \left(s\right) \rightarrow 3 C {a}^{2 +} \left(a q\right) + 2 P {O}_{4}^{3 -} \left(a q\right)$

The solubility constant is as standard, the concentrations of the ions raised to the power of their stoichiometric coefficients:

${K}_{s p} = {\left[P {O}_{4}^{3 -}\right]}^{2} {\left[C {a}^{2 +}\right]}^{3}$

Alternatively, consider the solubility of ferric sulfate:

$F {e}_{2} {\left(S {O}_{4}\right)}_{3} \left(s\right) \rightarrow 2 F {e}^{3 +} + 3 S {O}_{4}^{2 -}$

${K}_{s p} = {\left[F {e}^{3 +}\right]}^{2} {\left[S {O}_{4}^{2 -}\right]}^{3}$

This should not be too water soluble, but check this.

If this is not want you wanted I apologize.