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Chapter 16: Problem 31

A solution is \(0.035 M\) in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(0.035 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\). Solid \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is added without changing the volume of the solution. (a) Which salt, \(\mathrm{PbSO}_{4}\) or \(\mathrm{PbCrO}_{4}\), will precipitate first? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) when the salt in (a) first begins to precipitate?

Short Answer

Expert verified
Additionally, what is the concentration of Pb2+ when the salt in part (a) first begins to precipitate? Answer: PbSO4 will precipitate first, and the concentration of Pb2+ when PbSO4 first begins to precipitate is 4.57 x 10^{-7} M.

Step by step solution

01

Find the Ksp values of PbSO4 and PbCrO4

Look up the solubility product constants (Ksp) of PbSO4 and PbCrO4 in a textbook or online reference. The Ksp values are: PbSO4: Ksp = 1.6 x 10^{-8} PbCrO4: Ksp = 1.2 x 10^{-12}
02

Calculate Q for both salts and compare Q to Ksp

Ion product (Q) is calculated as follows: Q = [Pb2+][SO42-] for PbSO4 Q = [Pb2+][CrO42-] for PbCrO4 At the beginning, before any lead salt has precipitated, the concentrations of sulfate and chromate ions are equal to 0.035 M as given in the problem. Thus, we can write: Q(PbSO4) = [Pb2+](0.035) Q(PbCrO4) = [Pb2+](0.035) Divide each Q by their respective Ksp values: Q(PbSO4)/Ksp(PbSO4) = [Pb2+](0.035) / (1.6 x 10^{-8}) Q(PbCrO4)/Ksp(PbCrO4) = [Pb2+](0.035) / (1.2 x 10^{-12}) The salt for which the ratio Q/Ksp is smaller will precipitate first.
03

Determine which salt precipitates first

Comparing the Q/Ksp values for both salts, we find that: [Pb2+](0.035) / (1.6 x 10^{-8}) < [Pb2+](0.035) / (1.2 x 10^{-12}) So PbSO4 will precipitate first. Answer to part (a): PbSO4 will precipitate first.
04

Calculate the concentration of Pb2+ when PbSO4 begins to precipitate

Since PbSO4 is determined to precipitate first, we will use its Ksp value to find the concentration of Pb2+ when precipitation starts: Ksp(PbSO4) = 1.6 x 10^{-8} = [Pb2+][SO42-] Given that [SO42-] = 0.035 M, we can solve for [Pb2+]: [Pb2+] = Ksp(PbSO4) / [SO42-] [Pb2+] = (1.6 x 10^{-8}) / 0.035 [Pb2+] = 4.57 x 10^{-7} M Answer to part (b): The concentration of Pb2+ when PbSO4 first begins to precipitate is 4.57 x 10^{-7} M.

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Most popular questions from this chapter

One gram of \(\mathrm{PbCl}_{2}\) is dissolved in \(1.0 \mathrm{~L}\) of hot water. When the solution is cooled to \(25^{\circ} \mathrm{C}\), will some of the \(\mathrm{PbCl}_{2}\) crystallize out? If so, how much?

The concentrations of various cations in seawater, in moles per liter, are $$ \begin{array}{llllll} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} & \mathrm{Ca}^{2+} & \mathrm{A}^{3+} & \mathrm{Fe}^{3+} \\ \text { Molarity }(M) & 0.46 & 0.056 & 0.01 & 4 \times 10^{-7} & 2 \times 10^{-7} \\ \hline \end{array} $$ (a) At what \(\left[\mathrm{OH}^{-}\right]\) does \(\mathrm{Mg}(\mathrm{OH})_{2}\) start to precipitate? (b) At this concentration, will any of the other ions precipitate? (c) If enough \(\mathrm{OH}^{-}\) is added to precipitate \(50 \%\) of the \(\mathrm{Mg}^{2+}\), what percentage of each of the other ions will precipitate? (d) Under the conditions in (c), what mass of precipitate will be ob- tained from one liter of seawater?

Write net ionic equations for the reaction of \(\mathrm{H}^{+}\) with (a) \(\mathrm{Cu}_{2} \mathrm{~S}\) (b) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SrCO}_{3}\) (d) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Calculate the solubility (in grams per liter) of silver chloride in the following. (a) pure water (b) \(0.025 \mathrm{M} \mathrm{BaCl}_{2}\) (c) \(0.17 \mathrm{M} \mathrm{AgNO}_{3}\)

Which of the following statements are true? (a) For an insoluble metallic salt, \(K_{\text {sp }}\) is always less than 1 . (b) More \(\mathrm{PbCl}_{2}\) can be dissolved at \(100^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\). One can conclude that dissolving \(\mathrm{PbCl}_{2}\) is an exothermic process. (c) When strips of copper metal are added to a saturated solution of \(\mathrm{Cu}(\mathrm{OH})_{2}\), a precipitate of \(\mathrm{Cu}(\mathrm{OH})_{2}\) can be expected to form because of the common ion effect.
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