/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Would you expect that anions wou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Would you expect that anions would be physically closer to the oxygen or to the hydrogens of water molecules that surround it in solution?

Short Answer

Expert verified
Based on the electrostatic interactions, an anion would be physically closer to the hydrogens of water molecules that surround it in solution. This is because the anion is negatively charged, and the hydrogen atoms in water molecules carry a partial positive charge, attracting the anion towards them.

Step by step solution

01

Understanding of anion and water molecules

First, let's understand the components of the problem. An anion is a negatively charged ion, while a water molecule (H2O) consists of two hydrogen atoms covalently bonded to one oxygen atom. In a water molecule, oxygen is more electronegative than hydrogen, resulting in a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms.
02

Electrostatic interactions between anions and water molecules

Considering basic principles of electrostatic interactions, opposite charges attract each other, while like charges repel each other. In this case, since an anion is negatively charged, it will be attracted to the positively charged entities in its vicinity.
03

Position of anion in the water molecule

In a water molecule, the hydrogen atoms carry a partial positive charge, while the oxygen atom carries a partial negative charge. Since the anion is negatively charged, it will be attracted to the partial positive charges on the hydrogen atoms of the water molecules due to electrostatic interaction.
04

Conclusion

Based on the analysis of electrostatic interactions, we can conclude that an anion would be physically closer to the hydrogens of water molecules that surround it in solution, as it is attracted to their partial positive charges.

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 91Ó°ÊÓ!

Key Concepts

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

Anion Behavior in Solution
Anions, which are negatively charged ions, play a significant role when dissolved in a solution, particularly in water. When an anion is introduced to a water-based solution, it interacts with the water molecules in a very specific way. This behavior is influenced by their negative charge. As opposites attract in electrostatic interactions, anions will be drawn towards areas of positive charge. In water, the hydrogen atoms carry a partial positive charge due to the oxygen atom's higher electronegativity. Therefore, in a solution, an anion will tend to surround itself with the hydrogen ends of water molecules. By doing so, it minimizes the repulsion from the oxygen atom's negative charge, settling towards the stabilizing influence provided by the positively charged hydrogen atoms. This particular behavior of anions is foundationally explained by examining the dynamics of charge distribution in aqueous environments.
Water Molecule Polarity
To understand why anions prefer to interact with certain parts of a water molecule, it's crucial to grasp the concept of water molecule polarity. Water (Hâ‚‚O) is a polar molecule due to the difference in electronegativity between the oxygen and hydrogen atoms. Oxygen is much more electronegative compared to hydrogen, which means it pulls the shared electrons closer to itself.
This creates a partial negative charge on the oxygen atom, whereas the hydrogen atoms acquire a partial positive charge. The entire molecule, although neutral overall, displays this dipole nature with a slightly negative side and a slightly positive side.
  • The oxygen atom is electron-rich, creating a higher density of negative charge at that end.
  • The hydrogen atoms, pulled away from shared electrons, end up more positively charged.
This separation of charges within the molecule makes it highly effective in forming various intermolecular interactions, crucially with ions, like anions, in solution.
Charge Distribution
Understanding charge distribution helps clarify the behaviors of ionic interactions in solution. In the context of water molecules, charge distribution is uneven due to the unequal sharing of electrons in the hydrogen-oxygen bonds. This uneven charge distribution is primarily the result of the oxygen atom's electronegativity, which is significantly greater than that of the hydrogen atoms.
This results in a bent molecular structure, with the oxygen at the vertex and the hydrogens bonded at an angle. Such an orientation effectively separates negative and positive charges within the molecule, forming a dipole.
An anion, with its negative charge, experiences forces from these dipole interactions in water.
  • It is repelled by the oxygen's partial negative charge.
  • It is attracted to the partial positive charge of the hydrogen atoms.
This dynamic leads to strategic positioning of ions in solutions, which is critical for understanding the dissolution process, the solubility of substances, and the behavior of ions like anions in various solvent environments.

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

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-}\), (b) \(\mathrm{NO}_{3}^{-}\), (c) \(\mathrm{NH}_{4}^{+}\), (d) \(\mathrm{S}^{2-}\), (e) \(\mathrm{SO}_{4}^{2-} \cdot\) [Section 4.2]

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4}\), (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3}\), (c) strontium sulfide, \(\mathrm{SrS}\), (d) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\). (e) Locate sulfur in the periodic table in Exercise 4.47; what region is it in? (f) Which region(s) of the period table contains elements that can adopt both positive and negative oxidation numbers?

The following reactions (note that the arrows are pointing only one direction) can be used to prepare an activity series for the halogens: $$ \begin{gathered} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{gathered} $$ (a) Which elemental halogen would you predict is the most stable, upon mixing with other halides? (b) Predict whether a reaction will occur when elemental chlorine and potassium iodide are mixed. (c) Predict whether a reaction will occur when elemental bromine and lithium chloride are mixed.

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3}\) ? (b) If you take a \(10.0\)-mL portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L}\), what will be the concentration of the final solution?

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl}\), (b) \(\mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{CdSO}_{4}\).

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.