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Chapter 18: Problem 96

\(\mathrm{Cu}^{2+}\) and \(\mathrm{Pb}^{2+}\) are both present in an aqueous solution. To precipitate one of the ions and leave the other in solution, add (a) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) ;\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\); (c) \(\mathrm{HNO}_{3}(\mathrm{aq}) ;\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\).

Short Answer

Expert verified
The solution that will precipitate one of the ions and leave the other in solution is \(H_2 SO_4\).

Step by step solution

01

Consider the solubility rules

The first step is to consider the solubility rules in order to determine whether a particular salt will be soluble or not. According to these rules, sulfates (\(SO_4^{2-}\)) are generally soluble except with \(Pb^{2+}\), \(Ba^{2+}\), \(Hg^{2+}\), and \(Ag^{+}\). Similarly, most nitrate (\(NO_3^-\)) salts are soluble. Sulfides (\(S^{2-}\)) are typically insoluble except with group 1 elements, \(NH_4^+\), \(Ca^{2+}\), \(Sr^{2+}\), and \(Ba^{2+}\). Therefore, we can exclude (d) \(NH_4 NO_3\) as it will not precipitate either of the ions.
02

Test the remaining solutions

Now, we test the remaining solutions (a), (b) and (c) for either \(Cu^{2+}\) or \(Pb^{2+}\) to precipitate. (a) \(H_2 S\) will form \(CuS\) and \(PbS\). Both these sulfides are insoluble, meaning both \(Cu^{2+}\) and \(Pb^{2+}\) will precipitate. Therefore, (a) \(H_2 S\) does not serve our purpose. (b) \(H_2 SO_4\) will form \(CuSO_4\) and \(PbSO_4\). While \(CuSO_4\) is soluble, \(PbSO_4\) is insoluble, so it will precipitate \(Pb^{2+}\) leaving \(Cu^{2+}\) in solution. So solution (b) \(H_2 SO_4\) is correct. As a verification step, we can look at (c) \(HNO_3\), which will form \(Cu(NO_3)_2\) and \(Pb(NO_3)_2\). Both these nitrates are soluble, hence \(HNO_3\) will not precipitate either of the ions.
03

Summarize the Solution

From the analysis, \(H_2 SO_4\) is the solution which will precipitate \(Pb^{2+}\) and leave \(Cu^{2+}\) in solution whilst all other solutions do not meet the requirements. Thus, the correct solution is (b) \(H_2 SO_4\).

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Most popular questions from this chapter

A solution is saturated with magnesium palmitate \(\left[\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2}, \text { a component of bathtub ring }\right] \mathrm{at}\) \(50^{\circ} \mathrm{C} .\) How many milligrams of magnesium palmitate will precipitate from \(965 \mathrm{mL}\) of this solution when it is cooled to \(25^{\circ} \mathrm{C} ?\) For \(\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2},\) \(K_{\mathrm{sp}}=4.8 \times 10^{-12}\) at \(50^{\circ} \mathrm{C}\) and \(3.3 \times 10^{-12}\) at \(25^{\circ} \mathrm{C}\).

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?]

The slightly soluble solute \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is most soluble in (a) pure water; (b) \(0.10 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4} ;\) (c) \(0.25 \mathrm{M} \mathrm{KNO}_{3}\); (d) \(0.40 \mathrm{M} \mathrm{AgNO}_{3}\).

What is the solubility of \(\mathrm{MnS}\), in grams per liter, in a buffer solution that is \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}-0.500 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} ?\) For \(\mathrm{MnS}, K_{\mathrm{spa}}=3 \times 10^{7}\).

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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