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Chapter 5: Problem 27

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) \(\mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow\) (b) \(\mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow\) (c) \(\operatorname{FeS}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow\) (d) \(\mathrm{K}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow\) (e) \(\mathrm{Mg}(\mathrm{s})+\mathrm{H}^{+} \longrightarrow\)

Short Answer

Expert verified
(a) \( \mathrm{Ba}^{2+} + 2 \mathrm{OH}^{-} \longrightarrow \mathrm{Ba(OH)}_2 \) , (b) \( \mathrm{H}^{+} + \mathrm{Cl}^{-} \longrightarrow \mathrm{HCl} \) , (c) No reaction, (d) \( \mathrm{HCO}_{3}^{-} + \mathrm{H}^{+} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3} \) , (e) No reaction

Step by step solution

01

Analyzing and forming compound if possible (a)

In [A], Ba2+ attracts OH-, producing a precipitate of Ba(OH)2. Hence, the net ionic equation is: \( \mathrm{Ba}^{2+} + 2 \mathrm{OH}^{-} \longrightarrow \mathrm{Ba(OH)}_2 \)
02

Analyzing and forming compound if possible (b)

In [B], Cl- and H+ combine to form the compound HCl. Therefore, the net ionic equation is: \( \mathrm{H}^{+} + \mathrm{Cl}^{-} \longrightarrow \mathrm{HCl} \)
03

Analyzing and forming compound if possible (c)

In [C], the FeS does not combine with H+ and I-, thus there's no reaction occurring and no net ionic equation.
04

Analyzing and forming compound if possible (d)

In [D], bicarbonate ion [HCO3-] reacts with H+ to form a common weakly acidic compound, carbonic acid [H2CO3]. Hence, the net ionic equation is: \( \mathrm{HCO}_{3}^{-} + \mathrm{H}^{+} \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3} \)
05

Analyzing and forming compound if possible (e)

Rounding up in [E], a solid like Mg cannot form new compounds with ions unless it ionizes into Mg2+. This reaction is impossible unless in the right conditions, thus we have to say there's no reaction taking place.

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Most popular questions from this chapter

What volume of \(0.248 \mathrm{M} \mathrm{CaCl}_{2}\) must be added to \(335 \mathrm{mL}\) of \(0.186 \mathrm{M} \mathrm{KCl}\) to produce a solution with a concentration of \(0.250 \mathrm{M} \mathrm{Cl}^{-2}\) Assume that the solution volumes are additive.

Write a balanced equation for the redox reactions. (a) The reaction of aluminum metal with hydroiodic acid. (b) The reduction of vanadyl ion ( \(\mathrm{VO}^{2+}\) ) to vanadic ion \(\left(\mathrm{V}^{3+}\right)\) in acidic solution with zinc metal as the reducing agent. (c) The oxidation of methanol by chlorate ion in acidic solution, producing carbon dioxide gas, water, and chlorine dioxide gas as products.

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) $\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow$ (b) $\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow$

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)
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