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Chapter 18: Problem 64

Show that in qualitative cation analysis group \(1,\) if you obtain \(1.00 \mathrm{mL}\) of saturated \(\mathrm{PbCl}_{2}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C}\), sufficient \(\mathrm{Pb}^{2+}\) should be present to produce a precipitate of \(\mathrm{PbCrO}_{4}(\mathrm{s}) .\) Assume that you use \(1 \mathrm{drop}\) \((0.05 \mathrm{mL})\) of \(1.0 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) for the test.

Short Answer

Expert verified
Based on the calculated values for the concentrations of the ions involved and the calculated ion product (Q), if \(Q > Ksp\) for \(PbCrO_{4}\), then the precipitate of \(PbCrO_{4}(s)\) will indeed form.

Step by step solution

01

Write the Balanced Equation

Write down the balanced equation for the precipitation reaction. The formula would be \(Pb^{2+}(aq) + CrO_{4}^{2−}(aq) → PbCrO_{4}(s)\)
02

Calculate Initial Concentrations

Calculate the molar concentration of each ion. It can be derived using M1V1 = M2V2 where M1 is the initial molarity of K2CrO4, V1 is the initial volume of K2CrO4, M2 is the final molarity of Pb2+ and CrO4 2−, and V2 is the final volume of the mixture. From the given volume and molarity for the ions, calculate the initial concentration of \(Pb^{2+}\) and \(CrO_{4}^{2−}\). Hence, initial concentration of \(Pb^{2+}\) = (1 ml / 1.05 ml) * solubility of \(PbCl_{2}\) = \(x M\) and initial concentration of \(CrO_{4}^{2−}\) = (0.05 ml/ 1.05 ml) *1.0 M \(= y M\)
03

Calculate Ion Product and Compare with Ksp

Calculate the Ion product Q by multiplying the concentrations of the ions as per the stoichiometry of the balanced equation from step 1. Using the formula for the ion product, \(Q = [Pb^{2+}] * [CrO_{4}^{2−}]\), calculate Q. Compare Ion product (Q) with solubility product constant (Ksp) of \(PbCrO_{4}\). If \(Q > Ksp\), it indicates that the solution is supersaturated with ions and a precipitate will form. To prove the original exercise, Q should come out as greater than the Ksp.

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Most popular questions from this chapter

The addition of \(\mathrm{HCl}(\mathrm{aq})\) to a solution containing several different cations produces a white precipitate. The filtrate is removed and treated with \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) in 0.3 M HCl. No precipitate forms. Which of the following conclusions is (are) valid? Explain. (a) \(\mathrm{Ag}^{+}\) or \(\mathrm{Hg}_{2}^{2+}\) (or both) is probably present. (b) \(\mathrm{Mg}^{2+}\) is probably not present. (c) \(\mathrm{Pb}^{2+}\) is probably not present. (d) \(\mathrm{Fe}^{2+}\) is probably not present.

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)

Which of the following solids is (are) more soluble in a basic solution than in pure water: \(\mathrm{BaSO}_{4}, \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), \(\mathrm{Fe}(\mathrm{OH})_{3}, \mathrm{NaNO}_{3},\) or MnS? Explain.

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 \(\mathrm{MnS}\), \(K_{\mathrm{spa}}=3 \times 10^{7}\).

All but two of the following solutions yield a precipitate when the solution is also made \(2.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). Those two are (a) \(\mathrm{MgCl}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{FeCl}_{3}(\mathrm{aq})\); (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\); (e) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})\).
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