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Chapter 18: Problem 66

Write net ionic equations for the following qualitative cation analysis procedures. (a) precipitation of \(\mathrm{PbCl}_{2}(\mathrm{s})\) from a solution containing \(\mathrm{Pb}^{2+}\) (b) dissolution of \(\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})\) in a solution of \(\mathrm{NaOH}(\mathrm{aq})\) (c) dissolution of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) in \(\mathrm{HCl}(\mathrm{aq})\) (d) precipitation of \(\mathrm{CuS}(\mathrm{s})\) from an acidic solution of \(\mathrm{Cu}^{2+}\) and \(\mathrm{H}_{2} \mathrm{S}\)

Short Answer

Expert verified
The net ionic equations for these reactions are: (a) Pb2+ (aq) + 2Cl- (aq) -> PbCl2 (s), (b) Zn(OH)2 (s) + 2OH- (aq) -> Zn(OH)4^2- (aq), (c) Fe(OH)3 (s) + 3H+ (aq) -> Fe3+ (aq) + 3H2O (l) and (d) Cu2+ (aq) + S2- (aq) -> CuS (s).

Step by step solution

01

Write the chemical equation for precipitation of PbCl2

Pb2+ (aq) + 2Cl- (aq) -> PbCl2 (s). This reaction describes the formation of PbCl2 precipitate from the Pb2+ ion. The net ionic equation is thus the same as the full ionic equation in this case, because there are no spectator ions.
02

Write the chemical equation for dissolution of Zn(OH)2

Zn(OH)2 (s) + 2OH- (aq) -> Zn(OH)4^2- (aq). This reaction describes the solubility of Zn(OH)2. In this case, the net ionic equation and the full ionic equation are also the same, as there are no spectator ions.
03

Write the chemical equation for dissolution of Fe(OH)3 in HCl

Fe(OH)3 (s) + 3H+ (aq) -> Fe3+ (aq) + 3H2O (l). This reaction is the acid-base reaction between Fe(OH)3 and HCl. The net ionic equation in this case is the same as the full ionic equation, because there are no spectator ions.
04

Write the chemical equation for precipitation of CuS from an acidic solution

Cu2+ (aq) + S2- (aq) -> CuS (s). This reaction is the formation of a precipitate of CuS from Cu2+ and S2-. The net ionic equation is thus the same as the full ionic equation in this case, because there are no spectator ions.

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Most popular questions from this chapter

In the Mohr titration, \(\mathrm{Cl}^{-}(\mathrm{aq})\) is titrated with \(\mathrm{AgNO}_{3}(\text { aq })\) in solutions that are at about \(\mathrm{pH}=7\). Thus, it is suitable for determining the chloride ion content of drinking water. The indicator used in the titration is \(\mathrm{K}_{2} \mathrm{CrO}_{4}(\text { aq }) .\) A red-brown precipitate of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) forms after all the \(\mathrm{Cl}^{-}\) has precipitated. The titration reaction is \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s}) .\) At the equivalence point of the titration, the titration mixture consists of \(\mathrm{AgCl}(\mathrm{s})\) and a solution having neither \(\mathrm{Ag}^{+}\) nor \(\mathrm{Cl}^{-}\) in excess. Also, no \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) is present, but it forms immediately after the equivalence point. (a) How many milliliters of \(0.01000 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\) are required to titrate \(100.0 \mathrm{mL}\) of a municipal water sample having \(29.5 \mathrm{mg} \mathrm{Cl}^{-} / \mathrm{L} ?\) (b) What is \(\left[\mathrm{Ag}^{+}\right]\) at the equivalence point of the Mohr titration? (c) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) in the titration mixture to meet the requirement of no precipitation of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) until immediately after the equivalence point? (d) Describe the effect on the results of the titration if \(\left[\mathrm{CrO}_{4}^{2-}\right]\) were (1) greater than that calculated in part (c) or (2) less than that calculated? (e) Do you think the Mohr titration would work if the reactants were exchanged - that is, with \(\mathrm{Cl}^{-}(\text {aq })\) as the titrant and \(\mathrm{Ag}^{+}(\) aq) in the sample being analyzed? Explain.

Which of the following saturated aqueous solutions would have the highest \(\left[\mathrm{Mg}^{2+}\right]\): (a) \(\mathrm{MgCO}_{3} ;\) (b) \(\mathrm{MgF}_{2};\) (c) \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\) Explain.

Calculate \(\left[\mathrm{Cu}^{2+}\right]\) in a \(0.10 \mathrm{M} \mathrm{CuSO}_{4}(\) aq) solution that is also \(6.0 \mathrm{M}\) in free \(\mathrm{NH}_{3}\). \(\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})\) \(K_{\mathrm{f}}=1.1 \times 10^{13}\)

A handbook lists the \(K_{\mathrm{sp}}\) values \(1.1 \times 10^{-10}\) for \(\mathrm{BaSO}_{4}\) and \(5.1 \times 10^{-9}\) for \(\mathrm{BaCO}_{3} .\) When saturated \(\mathrm{BaSO}_{4}(\mathrm{aq})\) is also made with \(0.50 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}),\) a precipitate of \(\mathrm{BaCO}_{3}(\mathrm{s})\) forms. How do you account for this fact, given that \(\mathrm{BaCO}_{3}\) has a larger \(K_{\mathrm{sp}}\) than does \(\mathrm{BaSO}_{4} ?\)

Both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cu}^{2+}\) are present in the same aqueous solution. Which of the following reagents would work best in separating these ions, precipitating one and leaving the other in solution: \(\mathrm{NaOH}(\mathrm{aq}), \mathrm{HCl}(\mathrm{aq})\), \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}),\) or \(\mathrm{NH}_{3}(\mathrm{aq}) ?\) Explain your choice.
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