节: 共同效应 | 19.15 共同效应

章节大纲

That's quite a charge!
::这是相当高的收费!

Lithium carbonate is an essential component of lithium , which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding CO ₂ under high pressure to form the more soluble LiHCO ₃ . The is then depressurized to remove the carbon dioxide and the lithium carbonate precipitates out of solution.
::碳酸锂是锂的基本成分,通常比普通碱电池更耐用,从锂矿石中获取这种材料,方法是在高压下增加CO2,形成较易溶解的LiHCO3。然后进行减压,将二氧化碳和碳酸锂加速剂从溶液中去除。

Common Ion Effect
::共同效应

In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its in solution.
::在硫酸钙饱和溶液中,固体硫酸钙与其溶液之间存在平衡。

$\begin{aligned} {CaSO}_{4} (s) ⇄ {Ca}^{2 +} (a q) + {SO}_{4}^{2 -} (a q) K_{s p} = 2.4 \times 10^{- 5} \end{aligned}$
$\begin{align*}\text{CaSO}_4(s) \rightleftarrows \text{Ca}^{2+}(aq)+\text{SO}_4^{2-}(aq) \qquad K_{sp}=2.4 \times 10^{-5}\end{align*}$
::CASO4+Ca2+(aq)+SO42-(aq)Ksp=2.4×10-5

Suppose that some calcium nitrate were added to this saturated solution. Immediately, the of the calcium ion in the solution would increase. As a result, the ion product of the [Ca ²⁺ ] times the [SO ₄ ²⁻ ] would increase and now be greater than the $\begin{align*}K_{sp}\end{align*}$ . According to Le Châtelier’s principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the $\begin{align*}K_{sp}\end{align*}$ . Note that in the new equilibrium the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.
::假设在饱和溶液中添加了一些硝酸钙。溶液中的钙离子将立即增加。结果, [Ca2+]乘以[SO42-]的离子产物将增加,现在将大于Ksp。根据勒查特利尔的原则,以上平衡将转移到左边,以减轻添加的钙离子的压力。额外的硫酸钙将从溶液中产生,直到离子产物再次与Ksp等量。请注意,在新的平衡中,钙离子和硫酸盐离子的浓度将不再相互相等。钙离子浓度将大于硫酸离子的浓度。

This situation describes the common ion effect. A common ion is an ion that is in common to both salts in a solution. In the above example, the common ion is Ca ²⁺ . The common ion effect is a decrease in the of an as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional CaSO ₄ to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.
::共同离子是两种盐类在溶液中的共同离子。在以上例子中,共同离子是Ca2+。常见离子效应是增加一个共同离子的结果导致一种离子的减少。在硫酸钙的饱和溶液中添加钙离子会导致更多的CaSO4从溶液中解脱出来,降低溶液的溶性。添加含有硫酸盐离子的溶液,如硫酸钾,将产生同样的共同离子效应。

Sample Problem: The Common Ion Effect
::样本问题:共同效应

What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH has been added?
::氢氧化锌饱和溶液的1.00升中的锌离子浓度是多少?

Step 1: List the known quantities and plan the problem .
::第1步:列出已知数量并规划问题。

Known
::已知已知

$\begin{align*}K_{sp}= 3.0 \times 10^{-16}\end{align*}$ (from table in "Conversion of Ksp to Solubility" )
::Ksp=3. 0x10- 16 (取自“ Ksp 转换为溶性” 表格)

moles of added NaOH = 0.040 mol
::添加的 NaOH 摩尔 = 0.040 摩尔

volumes of solution = 1.00 L
::溶液容量=1.00升

Unknown
::未知

[Zn ²⁺ ] = ? M
::[Zn2+]=M?

Express the concentrations of the two ions relative to the variable $\begin{align*}s\end{align*}$ . The concentration of the zinc ion will be equal to $\begin{align*}s\end{align*}$ , while the concentration of the hydroxide ion will be equal to $\begin{align*}0.040 + 2s\end{align*}$ .
::锌离子的浓度等于S,而氢氧化离子的浓度等于0.040+2。

Step 2: Solve .
::步骤2:解决。

The $\begin{align*}K_{sp}\end{align*}$ expression can be written in terms of the variable $\begin{align*}s\end{align*}$ .
::Ksp 表达式可以用变量 s 写入。

$\begin{aligned} K_{s p} = [{Zn}^{2 +}] [{OH}^{-}]^{2} = (s) (0.040 + 2 s)^{2} \end{aligned}$
$\begin{align*}K_{sp}=[\text{Zn}^{2+}][\text{OH}^-]^2=(s)(0.040+2s)^2\end{align*}$
::Ksp=[Zn2+][OH-]2=(s)(0.040+2s)2

Because the value of the $\begin{align*}K_{sp}\end{align*}$ is so small, we can make the assumption that the value of $\begin{align*}s\end{align*}$ will be very small compared to 0.040. This simplifies the mathematics involved in solving for $\begin{align*}s\end{align*}$ .
::由于 Ksp 值如此之小, 我们可以假设 s 值与 0.040 相比将非常小。这简化了 s 的解决所涉及的数学。

$\begin{aligned} K_{s p} & = (s) (0.040)^{2} = 0.0016 s = 3.0 \times 10^{- 16} \\ s & = \frac{K_{s p}}{{[{OH}^{-}]}^{2}} = \frac{3.0 \times 10^{- 16}}{0.0016} = 1.9 \times 10^{- 13} M \end{aligned}$
$\begin{align*}K_{sp}&=(s)(0.040)^2=0.0016 \ s=3.0 \times 10^{-16} \\ s&=\frac{K_{sp}}{\left [\text{OH}^- \right ]^2}=\frac{3.0 \times 10^{-16}}{0.0016}=1.9 \times 10^{-13} \ \text{M}\end{align*}$
::Ksp=(s)(0.040)2=0.0016 s=3.0x10-16s=Ksp[OH-]2=3.0x10-160.0016=1.9x10-13 M

The concentration of the zinc ion is equal to $\begin{align*}s\end{align*}$ and so [Zn ²⁺ ] = 1.9 × 10 ^-13 M.
::锌离子的浓度等于s,因此[Zn2+]=1.9×10-13 M。

Step 3: Think about your result .
::步骤3:想想你的结果。

The relatively high concentration of the common ion, OH ⁻ , results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.
::普通离子(OH-)的相对高浓度导致锌离子的浓度非常低,氢氧化锌的摩尔溶性在存在普通离子时低于在水中的摩尔溶性。

Summary
::摘要

The common ion and common ion effect are described.
::介绍了共同离子效应和共同离子效应。

Calculations involving the common ion effect are described.
::说明涉及共同离子效应的计算。

Review
::回顾

How is Le Châtelier’s principle involved in the common-ion effect?
::勒查特利尔的原则如何涉及共同效应?

Could the common ion effect ever increase the solubility of a compound?
::共同离子效应是否会增加化合物的溶解性?

In the sample problem, what would the effect be of adding Zn(NO ₃ ) ₂ ?
::在抽样问题中,增加Zn(NO3)2的效果如何?

章节大纲

That's quite a charge! ::这是相当高的收费!

Common Ion Effect ::共同效应

Sample Problem: The Common Ion Effect ::样本问题:共同效应

Summary ::摘要

Review ::回顾

That's quite a charge!
::这是相当高的收费!

Common Ion Effect
::共同效应

Sample Problem: The Common Ion Effect
::样本问题:共同效应

Summary
::摘要

Review
::回顾