# Solubility Equilibria¶

• Molar Solubility ($$mol \cdot L^{−})$$
• Solubility $$g L^{−1}$$

$$K_{sp} =$$ solubility product

#### Written as equilibrium equation with $$K_{sp}$$ as reactants¶

$K_{sp} = \frac{[\ce{Ca}^{2+}]} {[\ce{F−}]^2}$

$$K_{sp}$$ values determine the solubility of a compound in water under specified conditions As $$K_{sp}$$ increases, so does solubility

$$K_{sp}$$ $$<<10^{-5}$$ $$10^{-5} \text{ to } 10^{-2}$$ $$>10^{-2}$$
Solubility Practically Insoluble Slightly Soluble Soluble

## Calculating the dissociation amount¶

1. Reaction
$\hskip{1cm}\ce{Cr(OH)3 <=> Cr^{3+} + 3OH−}$
1. ICE
$\begin{gather} Initial\hskip{2.3cm}0\hskip{1cm}0\\ Change\hskip{2.1cm}S\hskip{1cm}3S\\ Equilibrium\hskip{1.4cm}S\hskip{1cm}3S \end{gather}$
1. Substitute
\begin{align}K_{sp}&=\ce{[Cr^{3+}][OH−]^3}\\ K_{sp}&=(s)(3s)^3\\ K_{sp}&=(s)(27s^3)\\ K_{sp}&=27s^4\end{align}
1. Rearrange and solve for s (given stock $$K_{sp}$$ values)
\begin{align}K_{sp}&=3.0\times10^{−29}\\ \frac{K_{sp}}{27}&=s^4\\ \bigg[\frac{K_{sp}}{27}\bigg]^{1/4}&=s\\ \bigg[\frac{3.0×\times10^{−29}}{27}\bigg]^{1/4}&=s\\ 3.5\times10^{−8} M&=s\\ \end{align}

## Will Precipitation Occur¶

To determine if Precipitation will occur, first calculate $$Q_{sp}$$ (the $$K_{sp}$$ of the current conditions)

• If $$K_{sp}< Q_{sp}$$ Precipitate Wil Form - Too much in solution
• If $$K_{sp}> Q_{sp}$$ No Precipitation - Solution can hold more of the compound
• If $$K_{sp}= Q_{sp}$$ Already at equilibrium - No Change

## Converting these into masses¶

Determining the concentration of the solution and the volume of the solution will give us the number of moles

E.g. Determining #moles

$250\:mL \text{ of } 3.5\times10^{−8} \:M \text{ solution}$
\begin{align}n&=cv\\\ n&=(3.5\times10^{−8})\times 0.250\\ n&=8.8\times10^{−9}\:mol\end{align}

Determining mass:

\begin{align}\ce{Cr(OH)3}&=103.024\:au\\ m&=nM\\ m&=(8.8\times10^{−9})×\times 103.024\\ m&=9.1\times10^{−7} \:g\end{align}