Introduction to Solubility Equilibria
Introduced| English | Meaning |
|---|---|
| solubility product/ˌsɒljuːˈbɪlɪti ˈprɒdʌkt/ | The equilibrium constant for dissolution of a sparingly soluble ionic solid, expressed using the activities of its dissolved ions. |
| molar solubility/ˈməʊlə ˌsɒljuːˈbɪlɪti/ | The moles of a solid that dissolve per litre of saturated solution. |
How much will actually dissolve
- Even "insoluble" salts dissolve a tiny bit.
- A trickle of ions escapes into the water and no more.
- That trickle sets up its own equilibrium.
- A special constant measures exactly how much dissolves.
The solubility product
- The solubility product 溶度积 $K_{sp}$ is the equilibrium constant for dissolving.
- For $\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$, $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$.
- The solid itself is left out.
In a saturated solution dissolving and precipitating run at equal rates
For $\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$, the $K_{sp}$ expression is...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The ions multiply; the solid is left out.
The undissolved solid appears in the $K_{sp}$ expression.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Pure solids are omitted; only the ions appear.
Reading Ksp
- A larger $K_{sp}$ means a more soluble salt.
- A tiny $K_{sp}$ means very little dissolves.
- It is product-favoured only slightly, if at all.
For salts of the same type, a larger $K_{sp}$ means the salt is...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
A bigger $K_{sp}$ lets more dissolve.
From Ksp to solubility
- Molar solubility 摩尔溶解度 is how many moles dissolve per litre.
- Solve $K_{sp}$ for the ion concentrations to find it.
- More ions per formula unit change the math.
Solubility as an equilibrium
Sort each salt by whether its dissolving needs a solubility equilibrium (Ksp).
The number of moles that dissolve per litre is the ____ solubility.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Molar solubility $s$ is moles dissolved per litre.
For $\text{AgCl}$, $K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times s = s^2$.
- If $K_{sp} = 1.0\times10^{-10}$, then $s = \sqrt{K_{sp}} = 1.0\times10^{-5}\ \text{M}$.
- That is the molar solubility of $\text{AgCl}$.
For $\text{AgCl}$ with $K_{sp} = 4\times10^{-10}$, the molar solubility $s$? Enter the coefficient before $\times10^{-5}$ M.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$s = \sqrt{K_{sp}} = \sqrt{4\times10^{-10}} = 2\times10^{-5}\ \text{M}$.
For $\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$, the $K_{sp}$ in terms of solubility $s$ is...
Difficulty 1/5 · 100 coins + a small bonus on first completion today
$[\text{Ca}^{2+}] = s$, $[\text{F}^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$.
The solid is not in the $K_{sp}$ expression — only the dissolved ions. Watch the stoichiometry: for a 1:2 salt like $\text{CaF}_2$, $[\text{F}^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$. A bigger $K_{sp}$ means more soluble, but compare directly only for salts with the same ion ratio.
The solubility product $K_{sp}$ is the equilibrium constant for a salt dissolving, written from the ion concentrations (the solid is omitted). From it you find the molar solubility $s$ — for $\text{AgCl}$, $s = \sqrt{K_{sp}}$. Mind the stoichiometry for salts with more than one of an ion.