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Chapter 16: Problem 63

A solution is \(1 \times 10^{-4} M\) 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 \(K_{\mathrm{sp}}\) values are $K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},$ 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^2+ is added gradually to the solution containing NaF, Na2S, and Na3PO4 is: 1. Pb3(PO4)2 2. PbF2 3. PbS

Step by step solution

01

Calculate the concentration of anions in the solution

Given the concentration of NaF, Na2S, and Na3PO4 is 1 x 10^-4 M, the concentration of F^-, S^2-, and PO4^3- in the solution will also be 1 x 10^-4 M.
02

Calculate the saturation quotient (Q) for PbF2, PbS, and Pb3(PO4)2

Use the formula Q = [Pb^2+][A^n-]: For PbF2: Q = [Pb^2+][F^-] = [Pb^2+](1 x 10^-4 M) For PbS: Q = [Pb^2+][S^2-] = [Pb^2+](1 x 10^-4 M) For Pb3(PO4)2: Q = [Pb^2+]^3[PO4^3-]^2 = ([Pb^2+]^3)(1 x 10^-4 M)^2
03

Compare calculated Q values to the given Ksp values

The compound with the smallest Q value above its Ksp will precipitate first. We have: Ksp(PbF2) = 4 x 10^-8 = [Pb^2+](1 x 10^-4 M) Ksp(PbS) = 7 x 10^-29 = [Pb^2+](1 x 10^-4 M) Ksp(Pb3(PO4)2) = 1 x 10^-54 = ([Pb^2+]^3)(1 x 10^-4 M)^2 Calculate the [Pb^2+] values for which the Q values are equal to the Ksp values: [Pb^2+]_PbF2 = 4 x 10^-4 M [Pb^2+]_PbS = 7 x 10^-25 M [Pb^2+]_Pb3(PO4)2 = 1 x 10^-16 M
04

Determine the order of precipitation

Since the [Pb^2+] required to reach the Ksp is smallest for Pb3(PO4)2, it will precipitate first. The next smallest [Pb^2+] value corresponds to PbF2, and the largest to PbS. So, the order of precipitation is as follows: 1. Pb3(PO4)2 2. PbF2 3. PbS

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Most popular questions from this chapter

The concentration of \(\mathrm{Ag}^{+}\) in a solution saturated with \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) is $2.2 \times 10^{-4} \mathrm{M} .\( Calculate \)K_{\mathrm{sp}}\( for \)\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}.$

a. Calculate the molar solubility of \(\mathrm{AgBr}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) . b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of \(\mathrm{AgBr}\) will dissolve in \(250.0 \mathrm{mL}\) of $3.0 M\( \)\mathrm{NH}_{3}?$ e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

Consider a solution made by mixing \(500.0 \mathrm{mL}\) of $4.0 \mathrm{M} \mathrm{NH}_{3}\( and \)500.0 \mathrm{mL}\( of \)0.40 \mathrm{M} \mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\( reacts with \)\mathrm{NH}_{3}$ to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \qquad K_{1}=2.1 \times 10^{3}$$ $$\operatorname{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}$$ Determine the concentration of all species in solution.

Order the following solids (a–d) from least soluble to most soluble. Ignore any potential reactions of the ions with water. a. \(\mathrm{AgCl} \quad K_{s p}=1.6 \times 10^{-10}\) b. \(\mathrm{Ag}_{2} \mathrm{S} \quad K_{\mathrm{sp}}=1.6 \times 10^{-49}\) c. \(\mathrm{CaF}_{2} \quad K_{\mathrm{sp}}=4.0 \times 10^{-11}\) d. \(\mathrm{CuS} \quad K_{\mathrm{sp}}=8.5 \times 10^{-45}\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI?}\)
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