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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:

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