Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 30: Problem 3

a) Which atom(s) of the water molecule is closest to the cations in solution? Why? b) Which atom(s) of the water molecule is closest to the anions in solution? Why?

Short Answer

Expert verified
a) The Oxygen atom of the water molecule is the closest to the cations in solution because it carries a slight negative charge due to the polarity of the water molecule. b) The Hydrogen atoms of the water molecule are the closest to the anions in solution because they carry a slight positive charge due to the polarity of the water molecule.

Step by step solution

01

Identify the Polarity of Water molecule

Water (H2O) is a polar molecule. This is because oxygen is more electronegative than hydrogen, and therefore the shared electrons spend more time closer to oxygen than to hydrogen. As a result, the oxygen atom gets a partial negative charge, and the hydrogen atoms get a partial positive charge.
02

Edge close to cations

Cations are positively charged ions. In a solution, cations will be more attracted to atoms that carry a negative charge. In the water molecule, the oxygen atom carries a partial negative charge due to the polarity of the molecule. Therefore, in a solution, the oxygen atom of the water molecule will come closest to the cations due to attraction between opposite charges.
03

Edge close to anions

On the other hand, anions are negatively charged ions. In a solution, anions will be more attracted to atoms that carry positive charge. In the water molecule, the hydrogen atoms carry a partial positive charge due to the polarity. Therefore, in a solution, the hydrogen atoms of the water molecule will come closest to the anions due to attraction between opposite 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.

Start your free trial

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

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 is more concentrated with respect to sodium ions: \(50.0 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(500.0 \mathrm{~mL}\) of solution or \(59.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in \(500.0 \mathrm{~mL}\) of solution?

When \(1.000\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is dissolved in water to make exactly one liter of solution: a) how many moles of sodium ions are found in the solution? b) how many moles of sulfate ions are found in the solution? c) what is the molarity of sodium ions in the solution? d) what is the molarity of sulfate ions in the solution?

Draw a figure similar to the right side of Figure 1 for \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) in solution showing exactly 3 cations and the appropriate number of anions. (You can omit \(\mathrm{H}_{2} \mathrm{O}\) molecules for clarity. \()\)

Verify that when \(10.0 \mathrm{~g}\) of sodium sulfate dissolves in water: a) there are \(7.04 \times 10^{-2}\) moles of sodium sulfate in the water. b) there are \(7.04 \times 10^{-2}\) moles of sulfate in the water. c) there are \(14.1 \times 10^{-2}\) moles of sodium in the water.

Which is more concentrated with respect to sodium ions, \(0.30 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) or \(0.50 \mathrm{M} \mathrm{NaCl} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation
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