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Chapter 16: Problem 37

Write net ionic equations for the reaction of \(\mathrm{H}^{+}\) with (a) \(\mathrm{Cu}_{2} \mathrm{~S}\) (b) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SrCO}_{3}\) (d) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) (e) \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
Based on the given step-by-step solution, the net ionic equations for each reaction are: (a) Reaction between H⁺ and Cu₂S: 2H⁺ + Cu₂S → 2Cu⁺ + H₂S (b) Reaction between H⁺ and Hg₂Cl₂: Hg₂Cl₂ + 2H⁺ → Hg₂²⁺ + 2Cl⁻ + H₂ (c) Reaction between H⁺ and SrCO₃: SrCO₃ + 2H⁺ → Sr²⁺ + CO₂ + H₂O (d) Reaction between H⁺ and Cu(NH₃)₄²⁺: Cu(NH₃)₄²⁺ + 4H⁺ → Cu²⁺ + 2NH₄⁺ (e) Reaction between H⁺ and Ca(OH)₂: 2H⁺ + 2OH⁻ → 2H₂O

Step by step solution

01

(a) Reaction between \(\mathrm{H}^{+}\) and \(\mathrm{Cu}_{2} \mathrm{S}\)

Firstly, write the balanced chemical equation for the reaction: $$ \mathrm{Cu}_{2} \mathrm{S} + 2\mathrm{H}^{+} \longrightarrow 2\mathrm{Cu}^{+} + \mathrm{H}_{2} \mathrm{S} $$ Now, dissociate the reactants and products in their ionic form, considering their solubility in water: $$ \mathrm{Cu}_{2} \mathrm{S} + 2\mathrm{H}^{+} \longrightarrow 2\mathrm{Cu}^{+} + \mathrm{H}_{2} \mathrm{S} $$ Finally, omit spectator ions (which are common to both sides) to obtain the net ionic equation: $$ 2\mathrm{H}^{+} + \mathrm{Cu}_{2} \mathrm{S} \longrightarrow 2\mathrm{Cu}^{+} + \mathrm{H}_{2} \mathrm{S} $$
02

(b) Reaction between \(\mathrm{H}^{+}\) and \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)

Write the balanced chemical equation for the reaction: $$ \mathrm{Hg}_{2} \mathrm{Cl}_{2} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Hg}_{2}^{2+} + 2\mathrm{Cl}^{-} + \mathrm{H}_{2} $$ Dissociate the reactants and products in their ionic form, considering their solubility in water: $$ \mathrm{Hg}_{2} \mathrm{Cl}_{2} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Hg}_{2}^{2+} + 2\mathrm{Cl}^{-} + \mathrm{H}_{2} $$ The net ionic equation will be the same since there are no spectator ions: $$ \mathrm{Hg}_{2} \mathrm{Cl}_{2} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Hg}_{2}^{2+} + 2\mathrm{Cl}^{-} + \mathrm{H}_{2} $$
03

(c) Reaction between \(\mathrm{H}^{+}\) and \(\mathrm{SrCO}_{3}\)

Write the balanced chemical equation for the reaction: $$ \mathrm{SrCO}_{3} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Sr}^{2+} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} $$ Dissociate the reactants and products in their ionic form, considering their solubility in water: $$ \mathrm{SrCO}_{3} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Sr}^{2+} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} $$ The net ionic equation will be the same since there are no spectator ions: $$ \mathrm{SrCO}_{3} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Sr}^{2+} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} $$
04

(d) Reaction between \(\mathrm{H}^{+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\)

Write the balanced chemical equation for the reaction: $$ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+} + 4\mathrm{H}^{+} \longrightarrow \mathrm{Cu}^{2+} + 2\mathrm{NH}_{4}^{+} $$ Dissociate the reactants and products in their ionic form, considering their solubility in water: $$ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+} + 4\mathrm{H}^{+} \longrightarrow \mathrm{Cu}^{2+} + 2\mathrm{NH}_{4}^{+} $$ The net ionic equation will be the same since there are no spectator ions: $$ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+} + 4\mathrm{H}^{+} \longrightarrow \mathrm{Cu}^{2+} + 2\mathrm{NH}_{4}^{+} $$
05

(e) Reaction between \(\mathrm{H}^{+}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Write the balanced chemical equation for the reaction: $$ \mathrm{Ca}(\mathrm{OH})_{2} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Ca}^{2+} + 2\mathrm{H}_{2}\mathrm{O} $$ Dissociate the reactants and products in their ionic form, considering their solubility in water: $$ \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-} + 2\mathrm{H}^{+} \longrightarrow \mathrm{Ca}^{2+} + 2\mathrm{H}_{2}\mathrm{O} $$ Finally, cancel out the spectator ions (which are common to both sides) to obtain the net ionic equation: $$ 2\mathrm{H}^{+} + 2\mathrm{OH}^{-} \longrightarrow 2\mathrm{H}_{2}\mathrm{O} $$

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