Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The following compounds are water-soluble. What ions are produced by each compound in aqueous solution? (a) KI (c) \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) (b) \(\mathrm{Mg}\left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2}\) (d) NaCN

Short Answer

Expert verified
(a) K⁺ and I⁻, (b) Mg²⁺ and 2 CH₃CO₂⁻, (c) 2 K⁺ and HPO₄²⁻, (d) Na⁺ and CN⁻.

Step by step solution

01

Dissociation of KI

Potassium iodide (KI) is a salt composed of potassium (K) and iodide (I) ions. In an aqueous solution, it dissociates completely into its constituent ions. Therefore, the dissociation equation is: \[ \text{KI}_{(aq)} \rightarrow \text{K}^+_{(aq)} + \text{I}^-_{(aq)} \] This indicates that one mole of KI produces one mole of \(\text{K}^+\) ions and one mole of \(\text{I}^-\) ions.
02

Dissociation of K₂HPO₄

Potassium hydrogen phosphate (\(\mathrm{K}_{2} \mathrm{HPO}_{4}\)) is a salt formed from two potassium ions, one hydrogen ion, and one phosphate ion. In water, it dissociates as follows: \[ \text{K}_2\text{HPO}_4_{(aq)} \rightarrow 2\text{K}^+_{(aq)} + \text{HPO}_4^{2-}_{(aq)} \] The dissociation produces two \(\text{K}^+\) ions for each molecule and one \(\text{HPO}_4^{2-}\) ion.
03

Dissociation of Mg(CH₃CO₂)₂

Magnesium acetate \(\text{Mg(CH}_3\text{CO}_2)_2\) dissociates into magnesium ions and acetate ions in water. The dissociation equation is: \[ \text{Mg(CH}_3\text{CO}_2)_2_{(aq)} \rightarrow \text{Mg}^{2+}_{(aq)} + 2\text{CH}_3\text{CO}_2^-_{(aq)} \] Thus, one mole of magnesium acetate yields one mole of \(\text{Mg}^{2+}\) ions and two moles of \(\text{CH}_3\text{CO}_2^-\) ions.
04

Dissociation of NaCN

Sodium cyanide (NaCN) dissociates into sodium and cyanide ions in an aqueous solution. The dissociation equation is: \[ \text{NaCN}_{(aq)} \rightarrow \text{Na}^+_{(aq)} + \text{CN}^-_{(aq)} \] This means that one mole of NaCN produces one mole of \(\text{Na}^+\) ions and one mole of \(\text{CN}^-\) ions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Aqueous Solutions
In chemistry, an aqueous solution is simply a solution where the solvent is water. Water is a universal solvent, meaning it can dissolve a large variety of substances. This happens because water molecules are polar — they have positive and negative charges. Therefore, they can surround different ions or molecules and pull them apart, allowing them to "dissolve" in the water.
A key feature of aqueous solutions is their ability to facilitate chemical reactions by allowing ions to move freely. These solutions are crucial in many chemical processes, both industrial and biological. When substances dissolve in water, they can create an environment where new bonds form or where existing bonds are broken. This is why understanding aqueous solutions is essential for comprehending many principles in chemistry.
In our case, when compounds like KI, K₂HPO₄, Mg(CH₃CO₂)₂, and NaCN dissolve in water, they are said to dissociate, meaning they split into their component ions.
Ionic Compounds
Ionic compounds are chemical compounds composed of ions that are held together by electrostatic forces, often known as ionic bonds. These compounds are typically formed when metals, which lose electrons easily, interact with non-metals, which gain electrons easily. This exchange of electrons forms charged particles, or ions. Metals form positively charged ions (cations), and non-metals form negatively charged ions (anions).
When dissolved in an aqueous solution, ionic compounds dissociate into their constituent ions. For example:
  • Potassium iodide (KI) dissociates into K⁺ and I⁻ ions.
  • Potassium hydrogen phosphate (K₂HPO₄) yields K⁺ and HPO₄⁻ ions.
  • Magnesium acetate [Mg(CH₃CO₂)₂] forms Mg²⁺ and CH₃CO₂⁻ ions.
  • Sodium cyanide (NaCN) splits into Na⁺ and CN⁻ ions.
When you understand how ionic compounds behave in water, predicting the behavior of solutions becomes easier. Ionic compounds often have high melting and boiling points, and they conduct electricity when dissolved in water, thanks to their free-moving ions.
Chemical Equations
Chemical equations are symbolic representations of a chemical reaction where reactants are shown on the left, products on the right, and the arrow signifies the direction of the reaction. In the context of ionic dissociation in aqueous solutions, chemical equations are used to show how ionic compounds separate into individual ions.
For instance, for potassium iodide (KI), the chemical equation is:
\[ \text{KI}_{(aq)} \rightarrow \text{K}^+_{(aq)} + \text{I}^-_{(aq)} \]
This highlights how one formula unit of KI produces one potassium ion and one iodide ion in solution.
Chemical equations often maintain a balance between the number and type of each atom on both sides. They help chemists understand the quantitative aspects of chemistry, predicting the amount of product formed in a reaction or how much reactant is needed. Balanced equations are crucial for laboratory preparations and understanding natural processes.
By interpreting these equations for dissociation, you can grasp the fundamental concept of molecular and ionic interactions in solutions, providing a clear picture of what occurs change-wise when compounds dissolve in water.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Balance the following equations, and then classify each as a precipitation, acid-base, or gas-forming reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq})\)

Write two chemical equations, one that shows \(\mathrm{H}_{2} \mathrm{O}\) reacting (with HBr) as a Bronsted base and a second that shows \(\mathrm{H}_{2} \mathrm{O}\) reacting (with \(\mathrm{NH}_{3}\) ) as a Bronsted acid.

Write balanced chemical equations for the following reactions: (a) production of ammonia, \(\mathrm{NH}_{3}(\mathrm{g}),\) by combining \(\mathrm{N}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (b) production of methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\ell)\) by combining \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) (c) production of sulfuric acid by combining sulfur, oxygen, and water

Suppose you wish to prepare a sample of magnesium chloride. One way to do this is to use an acid-base reaction, the reaction of magnesium hydroxide with hydrochloric acid. $$\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ When the reaction is complete, evaporating the water will give solid magnesium chloride. Suggest at least one other way to prepare \(\mathrm{MgCl}_{2}\).

Which two of the following reactions are oxidationreduction reactions? Explain your answer briefly. Classify the remaining reaction. (a) \(\mathrm{CdCl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{CdS}(\mathrm{s})+2 \mathrm{NaCl}(\mathrm{aq})\) (b) \(2 \mathrm{Ca}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CaO}(\mathrm{s})\) (c) \(4 \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free