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For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) (e) CuBr.

Short Answer

Expert verified
The short answer for the net ionic equations for each slightly soluble salt reacting with a strong acid would be: (a) MnS (s) + 2 H⁺ (aq) → Mn²⁺ (aq) + H₂S (g) (b) PbF₂ (s) + 2 H⁺ (aq) → Pb²⁺ (aq) + 2 HF (aq) (c) No net ionic equation for AuCl₃ with strong acid. (d) Hg₂C₂O₄ (s) + 2 H⁺ (aq) → Hg₂²⁺ (aq) + H₂C₂O₄ (aq) (e) CuBr (s) + H⁺ (aq) → Cu²⁺ (aq) + HBr (aq)

Step by step solution

01

(a) MnS with strong acid

MnS is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Mn²⁺ ions and hydrogen sulfide (H₂S) gas. The balanced complete ionic equation is: MnS (s) + 2 HCl (aq) → Mn²⁺ (aq) + 2 Cl⁻ (aq) + H₂S (g) Removing the spectator ions (Cl⁻) yields the net ionic equation: MnS (s) + 2 H⁺ (aq) → Mn²⁺ (aq) + H₂S (g)
02

(b) PbF₂ with strong acid

PbF₂ is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Pb²⁺ ions and HF. The balanced complete ionic equation is: PbF₂ (s) + 2 HCl (aq) → Pb²⁺ (aq) + 2 Cl⁻ (aq) + 2 HF (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: PbF₂ (s) + 2 H⁺ (aq) → Pb²⁺ (aq) + 2 HF (aq)
03

(c) AuCl₃ with strong acid

AuCl₃ is soluble in water, so it won't react with strong acids. Therefore, there is no net ionic equation for this reaction.
04

(d) Hg₂C₂O₄ with strong acid

Hg₂C₂O₄ is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Hg₂²⁺ ions and oxalic acid (H₂C₂O₄). The balanced complete ionic equation is: Hg₂C₂O₄ (s) + 2 HCl (aq) → Hg₂²⁺ (aq) + 2 Cl⁻ (aq) + H₂C₂O₄ (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: Hg₂C₂O₄ (s) + 2 H⁺ (aq) → Hg₂²⁺ (aq) + H₂C₂O₄ (aq)
05

(e) CuBr with strong acid

CuBr is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Cu²⁺ ions and HBr. The balanced complete ionic equation is: CuBr (s) + HCl (aq) → Cu²⁺ (aq) + Cl⁻ (aq) + HBr (aq) Removing the spectator ions (Cl⁻) yields the net ionic equation: CuBr (s) + H⁺ (aq) → Cu²⁺ (aq) + HBr (aq)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slightly Soluble Salts
Slightly soluble salts are compounds that do not dissolve well in water. Instead of fully dissolving, only a small amount of the compound dissociates to produce ions.
Often, these salts have very low solubility in water, allowing just a tiny fraction of their molecules to break into ions.

This makes them right on the edge: not truly soluble, but not entirely insoluble either.
  • Their low solubility means the equilibrium heavily favors the solid state over the ionic state in a solution.
  • Common examples include MnS, PbF₂, Hg₂C₂O₄, and CuBr as seen in our exercise.
When these slightly soluble salts react with strong acids, the solubility can increase because the strong acid provides hydrogen ions (H⁺).
These ions can react with anions from the salt, forming molecules that might be gases or more soluble substances, thus helping to dissolve more of the salt.
  • For example, reacting MnS with HCl results in the formation of H₂S gas, removing sulfide ions from the solution and enabling more MnS to dissolve.
This behavior plays a critical role in balancing complete ionic and net ionic equations.
Strong Acids
Strong acids are characterized by their ability to completely dissociate into ions in solution. When these acids dissolve in water, they split entirely into hydrogen ions (H⁺) and their respective anions.
This 100% dissociation means that no molecules of the intact acid remain in solution.
  • Common strong acids include hydrochloric acid (HCl), sulfuric acid (H₂SO₄), and nitric acid (HNO₃).
  • In reactions, strong acids act as potent sources of H⁺ ions, particularly relevant when reacting with slightly soluble salts as they encourage ionization and dissolution.
In the example steps, HCl was used to dissociate the slightly soluble salts into their respective ions.
The supply of H⁺ provided by the strong acid allows the anions from the salts to form different products that are more stable for reaction or dissolution processes.

Understanding these reactions helps us derive net ionic equations by simplifying the compounds involved in the reactions and focusing only on the changes occurring.
Spectator Ions
Spectator ions are ions that exist in the solution but do not participate directly in the chemical reaction.
They appear unchanged on both sides of the chemical equation and essentially "watch" the reaction take place without getting involved.
  • These ions are crucial for maintaining charge balance in the overall solution but have no effect on the net ionic equation.
  • In the context of our exercise, chloride ions (Cl⁻) act as spectator ions in several reactions.
When writing net ionic equations, these spectator ions are removed to simplify and highlight the key chemical changes.
This step helps in focusing purely on the components of the reaction that undergo transformation, giving insight into the core process of the reaction.

By understanding which ions are spectators, students can simplify complex reactions and better grasp the fundamental chemistry involved.

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Most popular questions from this chapter

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(\mathrm{K}_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and \(K_{f}=1.1 \times 10^{5}\) for \(\mathrm{AgCl}_{2}^{-},\) calculate the equilibrium constant for the following reaction: $$ \mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) $$

A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(30.0 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 M\) HCl. Calculate the pH after the following volumes of titrant have been added: \((\mathbf{a}) 0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},\) (c) 59.0 \(\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .\)

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{+}\), (d) \(\mathrm{Ag}^{+}\) and \(\mathrm{Mn}^{2+} .\) Suggest how each mixture might be separated.

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