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A solution is \(1 \times 10^{-4} \mathrm{M}\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{~S}\), and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\text {sp }}\) values are \(K_{\text {sp }}\left(\mathrm{PbF}_{2}\right)=\) \(4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29}\), and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}\)

Short Answer

Expert verified
The order of precipitation when a source of Pb虏鈦 is added to the given solution would be: PbS 鈫 Pb鈧(PO鈧)鈧 鈫 PbF鈧. This is based on the Pb虏鈦 concentrations required for precipitation, which follow the order PbS < Pb鈧(PO鈧)鈧 < PbF鈧.

Step by step solution

01

Identify the ionic equations for the three compounds

In this situation, the solution initially contains NaF, Na2S, and Na3PO4; and, a source of Pb虏鈦 ions will be added. The potentially formed compounds are: PbF2, PbS, and Pb3(PO4)2. Write the ionic equations for the formation of these compounds: 1. Pb虏鈦 (aq) + 2F鈦 (aq) 鈫 PbF鈧 (s) 2. Pb虏鈦 (aq) + S虏鈦 (aq) 鈫 PbS (s) 3. 3Pb虏鈦 (aq) + 2PO鈧劼斥伝 (aq) 鈫 Pb鈧(PO鈧)鈧 (s)
02

Calculate the solubility of each compound

Using the Ksp values and the ionic equations, we can calculate the solubility of each compound in terms of Pb虏鈦 concentration: 1. PbF鈧: Ksp = [Pb虏鈦篯[F鈦籡虏 = \(4 \times 10^{-8}\) 鈬 [Pb虏鈦篯 = \(1 / (4 \times 10^{-8})\) since \(2 [F^{-}] = [Pb^{2+}]\) 2. PbS: Ksp = [Pb虏鈦篯[S虏鈦籡 = \(7 \times 10^{-29}\) 鈬 [Pb虏鈦篯 = \(7 \times 10^{-29}\) since \([S^{2-}] = [Pb^{2+}]\) 3. Pb鈧(PO鈧)鈧: Ksp = [Pb虏鈦篯鲁[PO鈧劼斥伝]虏 = \(1 \times 10^{-54}\) 鈬 [Pb虏鈦篯 = \(\sqrt[3]{1 / (1 \times 10^{-54})}\) = \(1 \times 10^{-18}\) since \((2/3) [PO_4^{3-}] = [Pb^{2+}]\)
03

Determine the order of precipitation

Compare the values of [Pb虏鈦篯 for all three compounds: 1. PbF鈧: [Pb虏鈦篯 = \(1 / (4 \times 10^{-8})\) 2. PbS: [Pb虏鈦篯 = \(7 \times 10^{-29}\) 3. Pb鈧(PO鈧)鈧: [Pb虏鈦篯 = \(1 \times 10^{-18}\) Since the Pb虏鈦 concentrations required for precipitation follow the order PbS < Pb鈧(PO鈧)鈧 < PbF鈧, the order of precipitation would be: PbS 鈫 Pb鈧(PO鈧)鈧 鈫 PbF鈧

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant (Ksp)
Understanding the solubility product constant, commonly abbreviated as Ksp, is essential when exploring the solubility of ionic compounds in aqueous solutions. It is a unique value for every sparingly soluble compound and represents the extent to which a compound can dissociate into its ions in solution. This constant is crucial for predicting whether a precipitation reaction will occur under specified conditions.

Let's take a closer look at what Ksp tells us. It is calculated as the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced chemical equation. For instance, the Ksp for a generic compound AB鈧 in water would be expressed as Ksp = [A鈦篯 x [B鈦籡虏, where [A鈦篯 and [B鈦籡 are the concentrations of the ions. In very simple terms, a lower Ksp value indicates a compound that is less soluble in water.

When lead (Pb虏鈦) is added to the solution containing NaF, Na鈧係, and Na鈧働O鈧, it's necessary to compare the Ksp values of the possible lead compounds to predict the order of precipitation. PbF鈧, PbS, and Pb鈧(PO鈧)鈧 will form and precipitate out based on the solubility limits defined by their Ksp values.
Ionic Compounds
Ionic compounds, such as salts, are composed of positive and negative ions held together by strong electrostatic forces known as ionic bonds. In the case of our exercise, we are dealing with compounds of lead (Pb虏鈦) combined with fluoride (F鈦), sulfide (S虏鈦), and phosphate (PO鈧劼斥伝) ions.

Ionic compounds typically dissolve in water by dissociating into their individual ions. The extent of their solubility varies and can be broadly predicted by the rules of solubility. However, Ksp provides a more precise and quantifiable approach, especially for compounds that are only sparingly soluble. Knowing the stoichiometry of these compounds, as well as their Ksp values, helps in predicting and explaining their behavior when they interact in a solution.

For example, in a given solubility equilibrium, high concentrations of certain ions can shift the balance, favoring the formation of a precipitate of the ionic compound. In our exercise, adding a source of Pb虏鈦 ions to a solution initially containing NaF, Na鈧係, and Na鈧働O鈧 will result in the formation of ionic compounds based on the solubility product constants.
Precipitation Reaction
A precipitation reaction occurs when two soluble salts react in solution to form one or more insoluble products, called precipitates. This type of reaction is common in chemistry and can be described using the ion product (Q), which is calculated in the same manner as the Ksp. However, Q represents the actual concentrations in the solution at a given moment, as opposed to the equilibrium concentrations described by the Ksp.

A precipitation reaction will take place if the ion product exceeds the Ksp for any given ionic compound. This is a straightforward way to predict the occurrence of precipitation 鈥 by comparing Q with Ksp. If Q > Ksp, precipitation is favored; if Q < Ksp, the compounds will remain in solution.

In our exercise, as Pb虏鈦 is gradually introduced to the solution, the ion product for PbS reaches its Ksp value first, followed by Pb鈧(PO鈧)鈧, and lastly PbF鈧. The first to reach this threshold will precipitate out of the solution, indicating the order in which the compounds will form precipitates: PbS precipitates first, followed by Pb鈧(PO鈧)鈧, and finally PbF鈧, in line with their respective solubility limits.

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Most popular questions from this chapter

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{-}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\).

Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is \(1.7 \times 10^{7} . K_{\text {sp }}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10}\). Calculate the molar solubility of AgCl in \(1.0 \mathrm{M} \mathrm{NH}_{3}\).

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}+\mathrm{P}_{3} \mathrm{O}_{10}^{5-} \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}{ }^{3-} .\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 . \mathrm{ppm} \mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L}\) of solution) after 40. g \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{~L}\) of the solution.

You are browsing through the Handbook of Hypothetical Chemistry when you come across a solid that is reported to have a \(K_{\text {sp }}\) value of zero in water at \(25^{\circ} \mathrm{C}\). What does this mean?

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