# Solubility Equilbria

Solubility Equilibria | Solubility Product.

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1 of 5 videos by Dr. Hayek

## Key Questions

Solubility equilibrium is a type of dynamic equiibrium.

#### Explanation:

It is important pharmaceutical, environmental and many other scenarios. This is because in pharmacy, you will need to measure the correct amount of substance to give patients appropriate medicine. In enviromental, the digging of salt or development of other pesticides must have the right components so that it does not cause harm.

By writing the appropriate solubility expression, and using the appropriate solubility constant, ${K}_{s p}$.

#### Explanation:

${K}_{s p}$ expressions are tabulated for a host of sparingly soluble and insoluble salts. They assume aqueous solubility and standard temperature and pressure.

Consider the solubility of a silver salt, say silver oxalate, in water at $298$ $K$. We could write the equilibrium this way:

$A {g}_{2} \left({C}_{2} {O}_{4}\right) \left(s\right) r i g h t \le f t h a r p \infty n s 2 A {g}^{+} \left(a q\right) + {C}_{2} {O}_{4}^{2 -} \left(a q\right)$.

As with any equilibrium we could write the express the equilibrium condition this way:

${K}_{s p} = \frac{{\left[A {g}^{+}\right]}^{2} \left[{C}_{2} {O}_{4}^{2 -}\right]}{\left[A {g}_{2} {C}_{2} {O}_{4} \left(s\right)\right]}$

${K}_{s p}$ will be small (sp stands for $\text{solubility product}$). And looking at the expression, we see that it contains $\left[A {g}_{2} {C}_{2} {O}_{4} \left(s\right)\right]$, the concentration of a solid, which is a meaningless concept, so can be removed.

Thus ${K}_{s p} = {\left[A {g}^{+}\right]}^{2} \left[{C}_{2} {O}_{4}^{2 -}\right]$.

Now ${K}_{s p}$ for silver oxalate is quoted as $3.6 \times {10}^{- 11}$ under standard conditions. (Of course I didn't know that off the top of my head.) ${K}_{s p}$'s have been measured and tabulated for a vast variety of insoluble and sparingly soluble salts. Why? Well, for one reason, if you are in the precious metal caper, and you precipitate out gold or platinum or rhodium salts, you had better make sure that the mother liquor (the solution from which the gold precipitates) is fairly free of gold content. You don't want to throw gold down the drain. If it were a cadmium, lead, or mercury salt, you also don't want to throw these metals down the drain for environmental reasons!

Back to the question; ${K}_{s p}$ for silver oxalate is quoted as $3.6 \times {10}^{- 11}$.

${K}_{s p} = {\left[A {g}^{+}\right]}^{2} \left[{C}_{2} {O}_{4}^{2 -}\right]$. If I call the solubility of silver oxalate, $S$. then ${K}_{s p} = 3.6 \times {10}^{- 11} = {\left(2 S\right)}^{2} \left(S\right) = 4 {S}^{3}$. So if I solve for $S$ $= \left\{\frac{1}{4} \sqrt[3]{3.6 \times {10}^{- 11}}\right\}$ I get an answer that represents the solubility of silver oxalate in water in $m o l \cdot {L}^{- 1}$.

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