In this chapter, we described an acid as a substance capable of producing \(\mathrm{H}^{+}\) and a salt as the ionic compound formed by the neutralization of an acid by a base. Write ionic equations to show that sodium hydrogen sulfate has the characteristics of both a salt and an acid (sometimes called an acid salt).

Short Answer

Expert verified
Sodium hydrogen sulfate \(\mathrm{NaHSO}_4\) can behave as both an acid and a salt. As an acid, it splits into \(\mathrm{H}^{+}\) and \(\mathrm{SO}_4^{2-}\) when in an aqueous solution: \(\mathrm{HSO}_4^{-} (aq) \rightarrow \mathrm{H}^{+} (aq) + \mathrm{SO}_4^{2-} (aq)\). As a salt, it can be formed through the neutralization of sulfuric acid (\(H_2SO_4\)) by sodium hydroxide (\(NaOH\)), as shown by: \(H_2SO_4 (aq) + 2NaOH (aq) \rightarrow 2H_2O (l) + NaHSO_4 (aq)\).

Step by step solution

01

Understanding the Molecular Structure

The molecular formula for sodium hydrogen sulfate is \(NaHSO_4\). It is made up of a sodium ion \(\mathrm{Na}^{+}\) and a hydrogen sulfate ion \(\mathrm{HSO}_4^{-}\).
02

Behavior as an Acid

As an acid, it should be capable of producing \(\mathrm{H}^{+}\) when dissolved in water. The ionic equation reflecting this is: \(\mathrm{HSO}_4^{-} (aq) \rightarrow \mathrm{H}^{+} (aq) + \mathrm{SO}_4^{2-} (aq)\). This equation shows that when hydrogen sulfate ion is in an aqueous solution, it splits into \(\mathrm{H}^{+}\) and \(\mathrm{SO}_4^{2-}\). This justifies it as an acid.
03

Behavior as a Salt

A salt is formed by the neutralization of an acid by a base. This can be demonstrated by showing the reaction of sulfuric acid (\(H_2SO_4\)) and sodium hydroxide (\(NaOH\)), a base. The ionic equation would be: \(H_2SO_4 (aq) + 2NaOH (aq) \rightarrow 2H_2O (l) + Na_2SO_4 (aq)\). If you remove one hydrogen from the reactant side, you end up with sodium hydrogen sulfate (\(NaHSO_4\)) instead of sodium sulfate (\(Na_2SO_4\)), showing it can be derived from a neutralization reaction.

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Most popular questions from this chapter

What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)

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Which of the following reactions are oxidationreduction reactions? (a) \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{CO}_{2}(\mathrm{g})\) (b) \(2 \mathrm{Li}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow 2 \mathrm{LiOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (c) \(4 \mathrm{Ag}(\mathrm{s})+\mathrm{PtCl}_{4}(\mathrm{aq}) \longrightarrow 4 \mathrm{AgCl}(\mathrm{s})+\mathrm{Pt}(\mathrm{s})\) (d) \(2 \mathrm{HClO}_{4}(\mathrm{aq})+\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(\mathrm{aq})\)

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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