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Chapter 18: Problem 28

Will \(\mathrm{PbCl}_{2}(\mathrm{s})\) precipitate when \(155 \mathrm{mL}\) of \(0.016 \mathrm{M}\) \(\mathrm{KCl}(\mathrm{aq})\) are added to \(245 \mathrm{mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) ?\)

Short Answer

Expert verified
To determine if \(\mathrm{PbCl}_{2}\) will precipitate, calculate the reaction quotient Q from the concentrations of Pb2+ and Cl- ions. If Q > Ksp of PbCl2, then \(\mathrm{PbCl}_{2}\) will precipitate.

Step by step solution

01

- Calculate the new concentration of Cl-

First of all, we need to find the concentration of Cl- ions in the mixed solution. Given that KCl is a strong electrolyte, every one molecule of KCl breaks down into its ions completely, so the Cl- concentration is the same as the KCl concentration. Since the total volume of the solution is (155mL + 245mL) = 400 mL, the new concentration of Cl- ions would be \((155 mL) \times(0.016 M) /(400 mL)\).
02

- Calculate the new concentration of Pb2+

In a similar manner, since Pb(NO3)2 is a strong electrolyte, the concentration of Pb2+ ions is the same as Pb(NO3)2 concentration. So the new concentration of Pb2+ ions would be \((245 mL) \times (0.175 M) /(400 mL)\).
03

- Calculate the reaction quotient (Q)

Now we calculate the reaction quotient (Q) which is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants, raised to their stoichiometric coefficients. Here, since the reaction forming the precipitate is \(\mathrm{Pb^{2+}}(\mathrm{aq})+\mathrm{2Cl^-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbCl}_2(s) \), we have Q = [Pb2+][Cl-]2. Substitute the calculated values for the concentrations of Pb2+ and Cl- ions into this formula to find the value of Q.
04

- Compare Q and Ksp of PbCl2

The solubility product constant (Ksp) of PbCl2 is \(1.7 \times 10^{-5}\). If the calculated Q is more than this Ksp, the PbCl2 will precipitate, as the solution is supersaturated. But if Q is less than or equal to Ksp, PbCl2 will not precipitate, as the solution can accommodate more PbCl2 without reaching saturation.

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Most popular questions from this chapter

Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

Determine the molar solubility of lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2},\) in a buffer solution with \(\mathrm{pH}=3.00,\) given that \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=2.5 \times 10^{-9}\) \(\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{a}}=1.9 \times 10^{-5}\)

If \(100.0 \mathrm{mL}\) of \(0.0025 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is saturated with \(\mathrm{CaSO}_{4},\) how many grams of \(\mathrm{CaSO}_{4}\) would be present in the solution? [Hint: Does the usual simplifying assumption hold?]

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.

An aqueous solution that \(2.00 \mathrm{M}\) in \(\mathrm{AgNO}_{3}\) is slowly added from a buret to an aqueous solution that is \(0.0100 \mathrm{M}\) in \(\mathrm{Cl}^{-}\) and \(0.250 \mathrm{M}\) in \(\mathrm{I}^{-}\). (a) Which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is the first to precipitate? (b) When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c) Is the separation of \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\) feasible by fractional precipitation in this solution?
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