19.12 从 Ksp 计算溶性

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  从 Ksp 计算溶性

  

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  How do you purify water?
  ::你们怎么清洁水呢?

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  Purification of water for drinking and other uses is a complicated process. Heavy need to be removed, a process accomplished by addition of carbonates and sulfates. Lead contamination can present major health problems, especially for younger children. Lead sulfates and carbonates are very insoluble, so will precipitate out of solution very easily.
  ::饮用水和其他用途用水的净化是一个复杂的过程,需要清除大量物质,通过添加碳酸盐和硫酸盐完成这一进程,铅污染可造成严重的健康问题,特别是对于幼儿,铅硫酸盐和碳酸盐非常难以溶解,因此很容易从溶液中解脱出来。

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  Conversion of  $\begin{align*}K_{sp}\end{align*}$ to Solubility
  ::Ksp 转换为溶性

  Product Constants (25°C)

  	$\begin{align*}K_{sp}\end{align*}$	Compound	$\begin{align*}K_{sp}\end{align*}$
  AgBr	5.0 × 10 -13	CuS	8.0 × 10 -37
  AgCl	1.8 × 10 -10	Fe(OH) 2	7.9 × 10 -16
  Al(OH) 3	3.0 × 10 -34	Mg(OH) 2	7.1 × 10 -12
  BaCO 3	5.0 × 10 -9	PbCl 2	1.7 × 10 -5
  BaSO 4	1.1 × 10 -10	PbCO 3	7.4 × 10 -14
  CaCO 3	4.5 × 10 -9	PbI 2	7.1 × 10 -9
  Ca(OH) 2	6.5 × 10 -6	PbSO 4	6.3 × 10 -7
  Ca 3 (PO 4 ) 2	1.2 × 10 -26	Zn(OH) 2	3.0 × 10 -16
  CaSO 4	2.4 × 10 -5	ZnS	3.0 × 10 -23

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  The known  $\begin{align*}K_{sp}\end{align*}$ values from  Table can be used to calculate the solubility of a given compound by following the steps listed below.
  ::表格中已知的 Ksp 值可用于计算特定化合物的溶性,具体方法如下:

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  1. Set up an ICE problem (Initial, Change, Equilibrium) in order to use the  $\begin{align*}K_{sp}\end{align*}$ value to calculate the of each of the .
     ::设置 ICE 问题( 初始、 变化、 平衡) , 以便使用 Ksp 值来计算每个 。
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  2. The concentration of the ions leads to the molar solubility of the compound.
     ::离子的浓度导致化合物的摩尔溶解性。
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  3. Use the to convert from molar solubility to solubility.
     ::用于从摩尔溶解性转换成溶解性。

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  The  $\begin{align*}K_{sp}\end{align*}$ of calcium carbonate is 4.5 × 10 ^-9 . We begin by setting up an ICE table showing the of CaCO ₃ into calcium ions and carbonate ions. The variable  $\begin{align*}s\end{align*}$ will be used to represent the molar solubility of CaCO ₃ . In this case, each formula unit of CaCO ₃ yields one Ca ²⁺ ion and one CO ₃ ²⁻ ion. Therefore, the equilibrium concentrations of each ion are equal to $\begin{align*}s\end{align*}$ .
  ::碳酸钙的 Ksp 碳酸钙为4.5 × 109. 我们首先建立一个ICE表格,显示CaCO3 的碳酸钙离子和碳酸盐离子。变量s将用来代表CaCO3的摩尔溶性。在这种情况下,CaCO3的每个公式单位产生1 Ca2+离子和1 CO32-离子。因此,每个离子的平衡浓度等于 S。

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  $$\begin{align*}& \text{CaCO}_3(s) \quad \rightleftarrows \quad \text{Ca}^{2+}(aq)+ \text{CO}_3^{2-}(aq) \\ \text{Initial }(\text{M}) & \qquad \qquad \qquad \qquad \quad 0.00 \qquad \quad \ 0.00 \\ \text{Change }(\text{M}) & \qquad \qquad \qquad \qquad \quad +s \qquad \quad +s \\ \qquad \text{Equilibrium }(\text{M}) & \qquad \qquad \qquad \qquad \qquad \ s \qquad \qquad s\end{align*}$$
  ::CaCO3+(aq)+(aq)+CO32-(aq)

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  The  $\begin{align*}K_{sp}\end{align*}$ expression can be written in terms of  $\begin{align*}s\end{align*}$ and then used to solve for $\begin{align*}s\end{align*}$ .
  ::Ksp 表达式可以用 s 写成, 然后用于 s 的解析 。

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  $$\begin{align*}K_{sp}&=[ \text{Ca}^{2+}][ \text{CO}_3^{2-}]=(s)(s)=s^2 \\ s&=\sqrt{K_{sp}}=\sqrt{4.5 \times 10^{-9}}=6.7 \times 10^{-5} \ \text{M}\end{align*}$$
  ::Ksp=[Ca2+][CO32-]=(s)(s)=s2s=Ksp=4.5×10-9=6.7×10-5M

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  The concentration of each of the ions at equilibrium is 6.7 × 10 ^-5  M. We can use the molar mass to convert from molar solubility to solubility.
  ::平衡时每种离子的浓度为6.7×10-5 M。 我们可以使用摩尔质量从摩尔溶性转换为溶性。

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  $$\begin{align*}\frac{6.7 \times 10^{-5} \ \cancel{\text{mol}}}{\text{L}} \times \frac{100.09 \ \text{g}}{1 \ \cancel{\text{mol}}} = 6.7 \times 10^{-3} \ \text{g/L}\end{align*}$$
  ::6.7x10-5 摩尔=100.09 g1 摩尔= 6.7x10-3克/升

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  So the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at 25°C is 6.7 × 10 ^-3  grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the  $\begin{align*}K_{sp}\end{align*}$ being equal to $\begin{align*}s^2\end{align*}$ . This is referred to as a formula of the type $\begin{align*}AB\end{align*}$ , where  $\begin{align*}A\end{align*}$ is the and  $\begin{align*}B\end{align*}$ is the . Now let’s consider a formula of the type $\begin{align*}AB_2\end{align*}$ , such as Fe(OH) ₂ . In this case the setup of the ICE table would look like the following:
  ::因此,25°C能溶解在1升水中的碳酸钙的最大碳酸钙含量为6.7×10-3克。请注意,在以上情况下,分离时离子的1:1比导致Ksp等于S2。这被称为AB型的公式,A是A,B是B。现在让我们考虑AB2型的公式,如Fe(OH)2。在这种情况下,ICE表格的设置将看起来如下:

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  $$\begin{align*}& \text{Fe(OH)}_2(s) \quad \rightleftarrows \quad \text{Fe}^{2+}(aq)+2 \text{OH}^-(aq) \\ \text{Initial }(\text{M}) & \qquad \qquad \qquad \qquad \qquad 0.00 \qquad \quad \ 0.00 \\ \text{Change }(\text{M}) & \qquad \qquad \qquad \qquad \qquad +s \qquad \quad +2s \\ \text{Equilibrium }(\text{M}) & \qquad \qquad \qquad \qquad \qquad \quad s \qquad \qquad \ 2s\end{align*}$$
  ::FE(OH)2(s) Fe2+(aq)+2OH-(aq) Initial(M)0.00 0.00 Change(M)+s+2s 平衡(M)s 2s

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  When the  $\begin{align*}K_{sp}\end{align*}$ expression is written in terms of $\begin{align*}s\end{align*}$ , we get the following result for the molar solubility.
  ::当 Ksp 表达式以 s 为单位写时, 我们得到以下的 molar 溶解性结果 。

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  $$\begin{align*}K_{sp}&=[ \text{Fe}^{2+}][ \text{OH}^-]^2=(s)(2s)^2=4s^3 \\ s&=\sqrt [3]{\frac{K_{sp}}{4}}=\sqrt [3]{\frac{7.9 \times 10^{-16}}{4}}=5.8 \times 10^{-6} \ \text{M}\end{align*}$$
  ::Ksp=[Fe2+][OH-]2=(s)(2s)2=4s3s=Ksp43=7.9×10-1643=5.8×10-6M

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  Table shows the relationship between  $\begin{align*}K_{sp}\end{align*}$ and molar solubility based on the formula.
  ::表格显示基于公式的 Ksp 和 molar 溶解性之间的关系。

  Compound Type	Example	$\begin{align*}K_{sp}\end{align*}$  Expression	Cation	Anion	$\begin{align*}K_{sp}\end{align*}$ in Terms of $\begin{align*}s\end{align*}$
  AB	CuS	[Cu 2+ ][S 2− ]	$\begin{align*}s\end{align*}$	$\begin{align*}s\end{align*}$	$\begin{align*}s^2\end{align*}$
  AB 2 or A 2 B	Ag 2 CrO 4	[Ag + ] 2 [CrO 4 2− ]	$\begin{align*}2s\end{align*}$	$\begin{align*}s\end{align*}$	$\begin{align*}4s^3\end{align*}$
  AB 3 or A 3 B	Al(OH) 3	[Al 3+ ][OH − ] 3	$\begin{align*}s\end{align*}$	$\begin{align*}3s\end{align*}$	$\begin{align*}27s^4 \end{align*}$
  A 2 B 3 or A 3 B 2	Ba 3 (PO 4 ) 2	[Ba 2+ ] 3 [PO 4 3− ] 2	$\begin{align*}3s\end{align*}$	$\begin{align*}2s\end{align*}$	$\begin{align*}108s^5 \end{align*}$

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  The  $\begin{align*}K_{sp}\end{align*}$ expressions in terms of  $\begin{align*}s\end{align*}$ can be used to solve problems in which the  $\begin{align*}K_{sp}\end{align*}$ is used to calculate the molar solubility as in the examples above. Molar solubility can then be converted to solubility.
  ::以 s 表示的 Ksp 表达式可以用来解决问题, Ksp 用来计算摩尔溶性, 如上例所示。 摩尔溶性可以转换成溶性 。

   

   

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  Summary
  ::摘要

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  • The process of determining solubilities using  $\begin{align*}K_{sp}\end{align*}$ values is described.
    ::介绍了使用 Ksp 值确定溶性的过程。

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  Review
  ::回顾

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  1. What information is needed to carry out these calculations?
     ::进行这些计算需要哪些信息?
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  2. What allows the calculation of molar solubility?
     ::什么允许计算摩尔溶性?
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  3. How is solubility determined?
     ::如何确定溶性?