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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鈦 (aq) 鈫 Mn虏鈦 (aq) + H鈧係 (g) (b) PbF鈧 (s) + 2 H鈦 (aq) 鈫 Pb虏鈦 (aq) + 2 HF (aq) (c) No net ionic equation for AuCl鈧 with strong acid. (d) Hg鈧侰鈧侽鈧 (s) + 2 H鈦 (aq) 鈫 Hg鈧偮测伜 (aq) + H鈧侰鈧侽鈧 (aq) (e) CuBr (s) + H鈦 (aq) 鈫 Cu虏鈦 (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 Mn虏鈦 ions and hydrogen sulfide (H鈧係) gas. The balanced complete ionic equation is: MnS (s) + 2 HCl (aq) 鈫 Mn虏鈦 (aq) + 2 Cl鈦 (aq) + H鈧係 (g) Removing the spectator ions (Cl鈦) yields the net ionic equation: MnS (s) + 2 H鈦 (aq) 鈫 Mn虏鈦 (aq) + H鈧係 (g)
02

(b) PbF鈧 with strong acid

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

(c) AuCl鈧 with strong acid

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

(d) Hg鈧侰鈧侽鈧 with strong acid

Hg鈧侰鈧侽鈧 is a slightly soluble salt, so when it reacts with the strong acid HCl, we expect to produce Hg鈧偮测伜 ions and oxalic acid (H鈧侰鈧侽鈧). The balanced complete ionic equation is: Hg鈧侰鈧侽鈧 (s) + 2 HCl (aq) 鈫 Hg鈧偮测伜 (aq) + 2 Cl鈦 (aq) + H鈧侰鈧侽鈧 (aq) Removing the spectator ions (Cl鈦) yields the net ionic equation: Hg鈧侰鈧侽鈧 (s) + 2 H鈦 (aq) 鈫 Hg鈧偮测伜 (aq) + H鈧侰鈧侽鈧 (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 Cu虏鈦 ions and HBr. The balanced complete ionic equation is: CuBr (s) + HCl (aq) 鈫 Cu虏鈦 (aq) + Cl鈦 (aq) + HBr (aq) Removing the spectator ions (Cl鈦) yields the net ionic equation: CuBr (s) + H鈦 (aq) 鈫 Cu虏鈦 (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, PbF鈧, Hg鈧侰鈧侽鈧, 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鈦).
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 H鈧係 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鈦) 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 (H鈧係O鈧), and nitric acid (HNO鈧).
  • In reactions, strong acids act as potent sources of H鈦 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鈦 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鈦) 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

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\).

How many milliliters of \(0.105 \mathrm{MHCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 \(\mathrm{mL}\) of \(0.0950 \mathrm{MNaOH}\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{MNH}_{3}\), (c) 125.0 mL of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

A buffer, consisting of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-},\) helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the \(\mathrm{pH}\) of a soft drink in which the major buffer ingredients are \(10.0 \mathrm{~g}\) of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) per \(0.500 \mathrm{~L}\) of solution?

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 -mL sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{MHCl}\) to reach the end point. Calculate \(K_{s p}\) for \(\mathrm{Ca}(\mathrm{OH})_{2} .\) Compare your result with that in Appendix D. Suggest a reason for any differences you find between your value and the one in Appendix D.

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