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Chapter 5: Problem 113

Classify each of the following statements as true or false. (a) Barium chloride, \(\mathrm{BaCl}_{2^{\prime}}\) is a weak electrolyte in aqueous solution. (b) In the reaction \(\mathrm{H}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{OH}^{-}(\mathrm{aq}),\) water acts as both an acid and an oxidizing agent. (c) A precipitate forms when aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}),\) is treated with excess aqueous hydrochloric acid, HCl(aq). (d) Hydrofluoric acid, \(\overline{\mathrm{HF}}\), is a strong acid in water. (e) Compared with a 0.010 M solution of \(\mathrm{NaNO}_{3}\), a \(0.010 \mathrm{M}\) solution of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is a better conductor of electricity.

Short Answer

Expert verified
Statement (a) is false, Statement (b) is false, Statement (c) is false, Statement (d) is false, Statement (e) is true.

Step by step solution

01

Statement (a) evaluation

BaCl2 is a salt of a strong acid (HCl) and a strong base (Ba(OH)2), so it completely ionises in aqueous solution. Therefore, it's a strong electrolyte, not a weak one. So, the statement is false.
02

Statement (b) evaluation

In this reaction, water is behaving as a base (accepting a proton from the hydride ion) and it's not involved in an electron transfer process to qualify as an oxidizing agent. Therefore, it's false to say that water is acting as both an acid and an oxidizing agent. So, the statement is false.
03

Statement (c) evaluation

When Na2CO3 reacts with HCl, it undergoes a double displacement reaction to produce NaCl and H2CO3. The H2CO3 then further decomposes into H2O and CO2 gas. So, there is no precipitate formed in this reaction. Therefore, the statement is false.
04

Statement (d) evaluation

HF, or hydrofluoric acid, is generally classified as a weak acid in water due to its relatively low level of ionization compared to the strong acids. So, the statement is false.
05

Statement (e) evaluation

A 0.010 M solution of Mg(NO3)2 will produce three times as many ions per liter as a 0.010 M solution of NaNO3. Because the conductivity of a solution is proportional to its ion concentration, the Mg(NO3)2 solution would be a better conductor of electricity. Thus, the statement is true.

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Most popular questions from this chapter

The titration of \(5.00 \mathrm{mL}\) of a saturated solution of sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\) requires \(25.8 \mathrm{mL}\) of \(0.02140 \mathrm{M} \mathrm{KMnO}_{4}\) in acidic solution. What mass of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in grams would be present in \(1.00 \mathrm{L}\) of this saturated solution? \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow_{\mathrm{Mn}^{2+}}+\mathrm{CO}_{2}(\mathrm{g}) \quad\) (not balanced)

Explain why these reactions cannot occur as written. (a) \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\)

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

Iron (Fe) is obtained from rock that is extracted from open pit mines and then crushed. The process used to obtain the pure metal from the crushed rock produces solid waste, called tailings, which are stored in disposal areas near the mines. The tailings pose a serious environmental risk because they contain sulfides, such as pyrite ( \(\mathrm{FeS}_{2}\) ), which oxidize in air to produce metal ions and \(\mathrm{H}^{+}\) ions that can enter into surface water or ground water. The oxidation of \(\mathrm{FeS}_{2}\) to \(\mathrm{Fe}^{3+}\) is described by the unbalanced chemical equation below. \(\mathrm{FeS}_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \quad(\text { not balanced })\) Thus, the oxidation of pyrite produces \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions that can leach into surface or ground water. The leaching of \(\mathrm{H}^{+}\) ions causes the water to become very acidic. To prevent acidification of nearby ground or surface water, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is added to the tailings to neutralize the \(\mathrm{H}^{+}\) ions: \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \underset{\mathrm{Ca}^{2+}}{\longrightarrow}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Balance the equation above for the reaction of \(\mathrm{FeS}_{2}\) and \(\mathrm{O}_{2}\). [ Hint: Start with the half-equations \(\mathrm{FeS}_{2}(\mathrm{s}) \rightarrow\) \(\left.\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \text { and } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) .\right]\) (b) What is the minimum amount of \(\mathrm{CaCO}_{3}(\mathrm{s})\) required, per kilogram of tailings, to prevent contamination if the tailings contain \(3 \%\) S by mass? Assume that all the sulfur in the tailings is in the form \(\mathrm{FeS}_{2}\).
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