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Chapter 18: Problem 42

A solution is \(0.010 \mathrm{M}\) in both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\). To this solution, \(0.50 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq })\) is slowly added. (a) Which anion will precipitate first from solution? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) at the point at which the second anion begins to precipitate? (c) Are the two anions effectively separated by this fractional precipitation?

Short Answer

Expert verified
a) The anion, \(\mathrm{CrO}_{4}^{2-}\) will precipitate first. b) The concentration of \(\left[\mathrm{Pb}^{2+}\right]\) is \(2.6 \times 10^{-6}\) M when the second anion begins to precipitate. c) Yes, the two anions are effectively separated by this fractional precipitation.

Step by step solution

01

determine the anion that precipitates first

To do this, use the solubility product constants of \(PbCrO_4\) and \(PbSO_4\), which are \(1.8 \times 10^{-14}\) and \(2.6 \times 10^{-8}\) respectively. The smaller solubility product constant belongs to \(PbCrO_4\), meaning it will precipitate first. This is because a smaller solubility product constant indicates a lower solubility in solution, and thus the compound will precipitate out of solution faster.
02

Find the concentration [Pb2+] at the start of the second anion's precipitation

Here, the solubility product constant of the second precipitate is used. Given that \(PbSO_4\) is the second to precipitate and its solubility product constant is \(K_{sp2} = 2.6 \times 10^{-8} = [Pb^{2+}][SO_4^{2-}]\), and given that the original concentration of \(SO_4^{2-}\) is 0.01M, the equation simplifies to \(2.6 \times 10^{-8} = [Pb^{2+}]\times (0.01)\). Solving for \([Pb^{2+}]\) gives \([Pb^{2+}] = 2.6 \times 10^{-6}\).
03

Determine if the two anions are effectively separated

If the concentration of \([Pb^{2+}]\) at the start of \(SO_4^{2-}\)'s precipitation is less than the initial \(CrO_4^{2-}\) concentration (that is \(2.6 \times 10^{-6} < 0.01\)), then complete precipitation of \(\mathrm{CrO}_{4}^{2-}\) occurs before \(\mathrm{SO}_{4}^{2-}\) begins to precipitate. This indicates that the two anions are effectively separated by fractional precipitation, which is indeed the case in this scenario.

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Most popular questions from this chapter

In a solution that is \(0.0500 \mathrm{M}\) in \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) and \(0.80 \mathrm{M}\) in free \(\mathrm{CN}^{-}\), the concentration of \(\mathrm{Cu}^{+}\) is \(6.1 \times 10^{-32} \mathrm{M}\) Calculate \(K_{\mathrm{f}}\) of \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\). \(\mathrm{Cu}^{+}(\mathrm{aq})+4 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}(\mathrm{aq}) \quad K_{\mathrm{f}}=?\)

A solution is \(0.05 \mathrm{M}\) in \(\mathrm{Cu}^{2+},\) in \(\mathrm{Hg}^{2+},\) and in \(\mathrm{Mn}^{2+}\). Which sulfides will precipitate if the solution is made to be \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) and \(0.010 \mathrm{M} \mathrm{HCl}(\mathrm{aq}) ?\) For \(\mathrm{CuS}\), \(K_{\mathrm{spa}}=6 \times 10^{-16} ;\) for \(\mathrm{HgS}, K_{\mathrm{spa}}=2 \times 10^{-32} ;\) for \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).

What percentage of the original \(\mathrm{Ag}^{+}\) remains in solution when \(175 \mathrm{mL} 0.0208 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(250 \mathrm{mL} 0.0380 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4} ?\)

Calculate the aqueous solubility, in moles per liter, of each of the following. (a) \(\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-10}\) (b) \(\mathrm{PbBr}_{2}, K_{\mathrm{sp}}=4.0 \times 10^{-5}\) (c) \(\mathrm{CeF}_{3}, K_{\mathrm{sp}}=8 \times 10^{-16}\) (d) \(\operatorname{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}, K_{\mathrm{sp}}=2.1 \times 10^{-20}\)

Appendix E describes a useful study aid known as concept mapping. Using the methods presented in Appendix \(\mathrm{E},\) construct a concept map that links the various factors affecting the solubility of slightly soluble solutes.
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