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The following compounds are water-soluble. What ions are produced by each compound in aqueous solution? (a) KOH (c) \(\mathrm{LiNO}_{3}\) (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) KOH: \(\text{K}^+\), \(\text{OH}^-\); (b) \(\text{K}_2\text{SO}_4\): \(\text{K}^+\), \(\text{SO}_4^{2-}\); (c) \(\text{LiNO}_3\): \(\text{Li}^+\), \(\text{NO}_3^-\); (d) \((\text{NH}_4)_2\text{SO}_4\): \(\text{NH}_4^+\), \(\text{SO}_4^{2-}\).

Step by step solution

01

Identify and Dissociate KOH

Potassium hydroxide (KOH) dissolves in water to produce potassium ions and hydroxide ions. The dissociation can be written as follows: \[\text{KOH} \rightarrow \text{K}^+ \text{(aq)} + \text{OH}^- \text{(aq)}\]. So the ions produced are \(\text{K}^+\) and \(\text{OH}^-\).
02

Dissociate \(\mathrm{LiNO}_{3}\)

Lithium nitrate (\(\mathrm{LiNO}_{3}\)) dissolves in water and dissociates into lithium ions and nitrate ions. The dissociation equation is: \[\text{LiNO}_3 \rightarrow \text{Li}^+ \text{(aq)} + \text{NO}_3^- \text{(aq)}\]. So the ions produced are \(\text{Li}^+\) and \(\text{NO}_3^-\).
03

Dissociate \(\mathrm{K}_{2} \mathrm{SO}_{4}\)

Potassium sulfate (\(\mathrm{K}_{2} \mathrm{SO}_{4}\)) in water dissociates into potassium ions and sulfate ions. The equation is: \[\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)}\]. Here, the ions produced are \(\text{K}^+\) and \(\text{SO}_4^{2-}\).
04

Dissociate \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\)

Ammonium sulfate (\((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\)) dissolves in water and dissociates into ammonium ions and sulfate ions. The dissociation is: \[(\text{NH}_4)_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)}\]. Therefore, the ions produced are \(\text{NH}_4^+\) and \(\text{SO}_4^{2-}\).

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Key Concepts

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

Ionic Compounds
Ionic compounds are substances composed of ions, which are atoms or molecules with a net electric charge due to the loss or gain of one or more electrons. These compounds are formed by a combination of positively charged ions (cations) and negatively charged ions (anions). A classic example of an ionic compound is sodium chloride, commonly known as table salt.
When ionic compounds are dissolved in water, they break apart into their constituent ions. This is because water is a polar solvent, and the positive and negative charges of the water molecules surround and stabilize the ions, pulling them apart from the solid structure of the ionic compound.
This dissociation process is essential for the conductivity of electricity in liquids, as ions are the carriers of electric charge in solutions. Students often come across the dissociation equations, such as for potassium hydroxide: \[ \text{KOH} \rightarrow \text{K}^+ \text{(aq)} + \text{OH}^- \text{(aq)} \] which highlights the formation of separate ions from the original compound.
Solubility in Water
Solubility refers to the ability of a substance to dissolve in a solvent, forming a homogeneous mixture. The solubility of ionic compounds in water varies widely; some dissolve readily, while others are insoluble. The dissolution process is influenced by the balance of forces between the ions of the compound and the polar water molecules.
Water, being a universal solvent, can dissolve many ionic compounds. However, solubility can be affected by several factors, including temperature, pressure, and the nature of both the solvent and solute.
When an ionic compound dissolves, it dissociates into its ions, as seen with lithium nitrate: \[ \text{LiNO}_3 \rightarrow \text{Li}^+ \text{(aq)} + \text{NO}_3^- \text{(aq)} \] In this equation, the water's ability to stabilize the lithium and nitrate ions facilitates their separation and dispersion into the solution.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. In the context of aqueous solutions, these reactions often involve reactants dissociating into ions and then reorganizing to form new products.
Dissociation is a key step in many chemical reactions involving ionic compounds. When potassium sulfate dissolves in water, it dissociates as follows: \[ \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \] This release of ions allows them to interact and participate in further chemical reactions, potentially forming precipitates, gases, or neutral molecules, depending on the reaction conditions and other reactants present. Understanding these steps is crucial in predicting the outcomes of chemical reactions and in balancing chemical equations.
Electrolytes
Electrolytes are substances that dissolve in water to produce solutions capable of conducting electricity. This property arises because electrolytes dissociate into ions, which can move freely and carry electric charge through the solution.
An example of an electrolyte is ammonium sulfate, which dissociates in water to produce ions that contribute to electrical conductivity: \[ (\text{NH}_4)_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \] Strong electrolytes, like many salts, dissociate completely in water, while weak electrolytes only partially dissociate. The degree of dissociation affects the strength of the electrolyte and its ability to conduct electricity.
Electrolytes play vital roles in biological systems, such as nerve impulse transmission and muscle contraction, as well as in everyday applications like battery operation and electrochemical cells.

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Most popular questions from this chapter

If you dilute \(25.0 \mathrm{mL}\) of \(1.50 \mathrm{M}\) hydrochloric acid to 500 . mL, what is the molar concentration of the dilute acid?

Balance the following equations, and then write the net ionic equation. Show states for all reactants and products \((s, \ell, g, a q)\) (a) the reaction of silver nitrate and potassium iodide to give silver iodide and potassium nitrate (b) the reaction of barium hydroxide and nitric acid to give barium nitrate and water (c) the reaction of sodium phosphate and nickel(II) nitrate to give nickel(11) phosphate and sodium nitrate

A Two students titrate different samples of the same solution of HCl using \(0.100 \mathrm{M} \mathrm{NaOH}\) solution and phenolphthalein indicator (see Figure 5.23). The first student pipets \(20.0 \mathrm{mL}\) of the HCl solution into a flask, adds 20 mL of distilled water and a few drops of phenolphthalein solution, and titrates until a lasting pink color appears. The second student pipets \(20.0 \mathrm{mL}\), of the HCl solution into a flask, adds 60 mL. of distilled water and a few drops of phenolphthalein solution, and titrates to the first lasting pink color. Each student correctly calculates the molarity of a HCl solution. What will the second student's result be? (a) four times less than the first student's result (b) four times greater than the first student's result (c) two times less than the first student's result (d) two times greater than the first student's result (e) the same as the first student's result

Balance the following reactions and then classify each as a precipitation, acid-base reaction, or a gas-forming reaction. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{KNO}_{3}(\mathrm{aq})\) (b) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq})\) (c) \(\mathrm{MgCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) \(\mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)

What feature causes the following reactions to be product-favored? (a) \(\mathrm{CuCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{CuS}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq})\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+3 \mathrm{KOH}(\mathrm{aq}) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{K}_{3} \mathrm{PO}_{4}(\mathrm{aq})\)

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