19.13 从溶性计算 Ksp 的计算

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  从溶性计算 Ksp 的计算

  

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  How is baking soda made?
  ::烤苏打如何做?

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  Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed which then reacts with the NaCl to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.
  ::碳酸铵首先形成,然后与纳氏反应,形成碳酸钠和氯化铵。 碳酸钠与其他材料相比不易溶解,因此会产生溶解。

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

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  is normally expressed in g/L of saturated solution. However, solubility can also be expressed as the moles per liter . Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given represents the highest solution that is possible for that compound. The of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of Zn(OH) ₂ is 4.2 × 10 ^-4  g/L, the molar solubility can be calculated as shown below:
  ::溶解度通常以饱和溶液的克/升表示。但是,溶解度也可以以每升的摩尔表示。摩尔溶性是指在饱和溶液的一升内溶液溶液的摩尔数。换句话说,给定溶液的摩尔溶性是该化合物可能达到的最高溶液。化合物是溶解性和摩尔溶性之间的转换系数。鉴于Zn(OH)2的溶解性为4.2 x 10-4克/升,可计算摩尔溶性如下:

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  $$\begin{align*}\frac{4.2 \times 10^{-4} \ \cancel{\text{g}}}{\text{L}} \times \frac{1 \ \text{mol}}{99.41 \ \cancel{\text{g}}}=4.2 \times 10^{-6} \ \text{mol/L} \ (\text{M})\end{align*}$$
  ::4.2x10-4 gL1 mol99.41 g=4.2x10-6 mol/L(M)

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  Solubility data can be used to calculate the  $\begin{align*}K_{sp}\end{align*}$ for a given compound. The following steps need to be taken.
  ::溶性数据可用于计算某一化合物的 Ksp 。 需要采取以下步骤。

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  1. Convert from solubility to molar solubility.
     ::从溶解度转换为摩尔溶解度。
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  2. Use the equation to determine the of each of the in mol/L.
     ::使用方程来确定摩尔/升中的每个值。
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  3. Apply the $\begin{align*}K_{sp}\end{align*}$ equation.
     ::应用 Ksp 方程式 。

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  Sample Problem: Calculating  $\begin{align*}K_{sp} \end{align*}$ from Solubility
  ::样本问题: 从溶性计算 Ksp

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  The solubility of lead(II) fluoride is found experimentally to be 0.533 g/L. Calculate the  $\begin{align*}K_{sp}\end{align*}$ for lead(II) fluoride.
  ::铅(II)氟化物的溶解性经实验认定为0.533克/升。 计算铅(II)氟化物的Ksp。

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

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  Known
  ::已知已知

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  • solubility of PbF ₂ = 0.533 g/L
    ::PbF2 = 0.533克/升的溶性
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  • molar mass = 245.20 g/mol
    ::摩尔质量=245.20克/摩尔

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  Unknown
  ::未知

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  • $\begin{align*}K_{sp}\end{align*}$ of PbF ₂ = ?
    ::{\fn方正准圆简体\fs12\1cHC9ECC4}PbF2 Ksp of PbF2 =? {\fn方正准圆简体\fs12\1cHC9ECC4}?

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  The dissociation equation for PbF ₂ and the corresponding  $\begin{align*}K_{sp}\end{align*}$ expression
  ::PbF2 和相应的 Ksp 表达式的分离方程式

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  $$\begin{align*}\text{PbF}_2(s) \rightleftarrows \text{Pb}^{2+}(aq)+2\text{F}^-(aq) && K_{sp}=[\text{Pb}^{2+}][\text{F}^-]^2 \end{align*}$$
  ::PbF2(s)Pb2+(aq)+2F(aq)Ksp=[Pb2+][F-]2

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  The steps above will be followed to calculate the  $\begin{align*}K_{sp}\end{align*}$ for PbF ₂ .
  ::将遵循上述步骤计算 PbF2 的 Ksp 。

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  Step 2: Solve .
  ::步骤2:解决。

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  $$\begin{align*}\text{molar solubility} \qquad \frac{0.533 \ \cancel{\text{g}}}{\text{L}} \times \frac{1 \ \text{mol}}{245.20 \ \cancel{\text{g}}}=2.17 \times 10^{-3} \ \text{M}\end{align*}$$
  ::摩勒溶溶溶性0.533 gLx1 mol245.20 g=2.17×10-3 M

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  The dissociation equation shows that for every mole of PbF ₂ that dissociates, 1 mol of Pb ²⁺ and 2 mol of F ⁻ are produced. Therefore, at equilibrium the concentrations of the ions are:
  ::分离方程式表明,对于每一个分离的PbF2摩尔,均产生1毫升的Pb2+和2毫升的F。

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  $$\begin{align*}[\text{Pb}^{2+}]=2.17 \times 10^{-3} \ \text{M} \quad \text{and} \quad [\text{F}^-]=2 \times 2.17 \times 10^{-3}=4.35 \times 10^{-3} \ \text{M} \end{align*}$$
  ::[Pb2+]=2.17×10-3毫米[F-]=2×2.17×10-3毫米[F-]=2×2.17×10-3=4.35×10-3米]

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  Substitute into the expression and solve for the $\begin{align*}K_{sp}\end{align*}$ .
  ::取代 Ksp 的表达式和解决方案 。

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  $$\begin{align*}K_{sp}=(2.17 \times 10^{-3})(4.35 \times 10^{-3})^2=4.11 \times 10^{-8}\end{align*}$$
  ::Ksp= (2. 17x10- 3)( 4. 35x10-3) 2= 4. 11x10-8

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

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  The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF ₂ .
  ::对于PbF2等几乎不可溶解的化合物而言,溶解性产品常数明显小于1。

   

   

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  Summary
  ::摘要

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  • Molar solubility calculations are described.
    ::介绍了摩尔溶性计算。
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  • Calculations of  $\begin{align*}K_{sp}\end{align*}$ using molar solubility are described.
    ::介绍了使用摩尔溶解度计算Ksp的计算方法。

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  Review
  ::回顾

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  1. What are the solution requirements for determining molar solubility?
     ::确定摩尔溶性的解决办法要求是什么?
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  2. Why do we need to convert mass to molarity to determine $\begin{align*}K_{sp}\end{align*}$ ?
     ::为什么我们需要将质量转换成质度 才能确定Ksp?
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  3. What  $\begin{align*}K_{sp}\end{align*}$ values would you expect for very insoluble compounds?
     ::您对非常不可溶解的化合物期望 Ksp 值是什么 ?