Introduction to Solubility Equilibria · 溶解平衡简介
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| solubility product/ˌsɒljuːˈbɪlɪti ˈprɒdʌkt/ | 溶度积 | róng dù jī |
| molar solubility/ˈməʊlə ˌsɒljuːˈbɪlɪti/ | 摩尔溶解度 | mó ěr róng jiě dù |
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.
究竟能溶解多少
- 即使是"不溶"的盐也会溶解一点点。
- 一小股离子逃进水里,再无更多。
- 那一小股建立起它自己的平衡。
- 一个特殊的常数精确测量能溶解多少。
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.
溶度积
- 溶度积 $K_{sp}$ 是溶解的平衡常数。
- 对 $\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$,$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$。
- 固体本身不列入。
For · 支持 $\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$, the $K_{sp}$ expression is... · 对于$\text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$，$K_{sp}$表达式为...
The ions multiply; the solid is left out. · 离子相乘；固体被省略。
The undissolved solid appears in the $K_{sp}$ expression. · 未溶解的固体出现在$K_{sp}$表达式中。
Pure solids are omitted; only the ions appear. · 纯固体被省略；仅离子出现。
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.
读 Ksp
- 更大的 $K_{sp}$ 意味着更易溶的盐。
- 微小的 $K_{sp}$ 意味着溶解得很少。
- 它即使偏向产物也只是稍微。
For salts of the same type, a larger $K_{sp}$ means the salt is... · 对于同类型的盐，较大的$K_{sp}$意味着该盐...
A bigger $K_{sp}$ lets more dissolve. · 更大的 $K_{sp}$ 允许溶解更多物质。
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.
从 Ksp 到溶解度
- 摩尔溶解度是每升溶解多少摩尔。
- 解 $K_{sp}$ 求出离子浓度即可得到它。
- 每个化学式单元的离子越多,算法就不同。
Solubility as an equilibrium · 作为平衡的溶解度
Sort each salt by whether its dissolving needs a solubility equilibrium (Ksp). · 按是否需溶解度平衡(Ksp)对每种盐排序。
The number of moles that dissolve per litre is the ____ solubility. · 每升溶解的摩尔数是____溶解度。
Molar solubility $s$ is moles dissolved per litre. · 摩尔溶解度$s$是每升溶解的摩尔数。
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 $\text{AgCl}$.
对 $\text{AgCl}$,$K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times s = s^2$。
- 若 $K_{sp} = 1.0\times10^{-10}$,则 $s = \sqrt{K_{sp}} = 1.0\times10^{-5}\ \text{M}$。
- 那就是 $\text{AgCl}$ 的摩尔溶解度。
For · 支持 $\text{AgCl}$ with $K_{sp} = 4\times10^{-10}$, the molar solubility $s$? Enter the coefficient before $\times10^{-5}$ M. · 对于 $\text{AgCl}$ 与 $K_{sp} = 4\times10^{-10}$，摩尔溶解度 $s$？请输入 $\times10^{-5}$ M 前的系数。
$s = \sqrt{K_{sp}} = \sqrt{4\times10^{-10}} = 2\times10^{-5}\ \text{M}$.
For · 支持 $\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$, the $K_{sp}$ in terms of solubility $s$ is... · 对于 $\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$， $K_{sp}$ 在溶解度方面 $s$ 是...
$[\text{Ca}^{2+}] = s$, $[\text{F}^-] = 2s$, so $K_{sp} = s(2s)^2 = 4s^3$. · $[\text{Ca}^{2+}] = s$，$[\text{F}^-] = 2s$，因此 $K_{sp} = s(2s)^2 = 4s^3$。
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.
固体不在 $K_{sp}$ 表达式里——只有溶解的离子。注意化学计量:对像 $\text{CaF}_2$ 这样的 1:2 盐,$[\text{F}^-] = 2s$,所以 $K_{sp} = s(2s)^2 = 4s^3$。更大的 $K_{sp}$ 意味着更易溶,但只对离子比相同的盐才能直接比较。
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.
溶度积 $K_{sp}$ 是盐溶解的平衡常数,由离子浓度写出(固体略去)。由它可求出摩尔溶解度 $s$——对 $\text{AgCl}$,$s = \sqrt{K_{sp}}$。对含有多个某离子的盐,要注意化学计量。