Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 7

Directions: Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. A stock solution of \(2.0 \mathrm{M} \mathrm{MgCl}_{2}\) is dissolved in water. (a) (i) In the beaker below, draw a particulate diagram that represents \(\mathrm{MgCl}_{2}\) dissolved in water. The approximate sizes of each atom/ion are provided for you. Your diagram should include at least four water molecules, which should be correctly oriented compared to the ions dissolved in solution. (DIAGRAM CANT COPY) (ii) Why are the chloride ions from (a)(i) larger than the magnesium (b) (i) A student wishes to make up 500 \(\mathrm{mL}\) of 0.50 \(M \mathrm{MgCl}_{2}\) for an experiment. Explain the best method of doing so utilizing a graduated cylinder and a volumetric flask. Assume \(\mathrm{MgCl}_{2}\) is fully soluble. (ii) What are the concentrations of the \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions in the new solution?

Short Answer

Expert verified
In the particulate diagram, Mg2+ ions are surrounded by the negative ends of the water (H2O) molecules, while Cl- ions are surrounded by the positive ends of water molecules. Cl- ions are larger due to more electron shells. The student should take 125 mL of a 2.0 M MgCl2 stock solution and dilute it into 500 mL to get a 0.50 M MgCl2 solution. The resulting solution has 0.50 M concentration of Mg2+ ions and 1.0 M of Cl- ions.

Step by step solution

01

Drawing MgCl2 Particulate Diagram

Draw MgCl2 molecule surrounded by 4 or more water molecules. The charge on Mg should be +2 (as it loses 2 electrons in the solution to form Mg2+ ions), and each Cl should be -1 (as they gain 1 electron each from Mg2+ to form Cl- ions). When hydrated, the negatively charged oxide ends of water molecules are attracted to and surround Mg2+ ions, and the positively charged hydrogen ends of the water molecules do the same for the Cl- ions.
02

Identifying the Sizes of the Ions

The chloride ions are larger than the magnesium because the chloride ion has more electron shells than the magnesium ion. These additional shells make the ion radius of chloride larger.
03

Diluting the stock solution

First, determined the volume of stock solution needed. Using \(M_{1}V_{1} = M_{2}V_{2}\) where \(M_{1}\) = 2 M, \(V_{1}\) is what we're looking for, \(M_{2}\) = 0.5 M, and \(V_{2}\) = 500 mL, you find \(V_{1}\) to be 125 mL. So, the student should measure and take 125 mL of the 2.0 M MgCl2 stock solution using a graduated cylinder and then transfer this solution into a 500 mL of volumetric flask. The flask should then be filled up to the 500 mL mark with distilled water to make a 0.50 M MgCl2 solution.
04

Determining the New Concentrations

As MgCl2 dissociates into one Mg2+ ion and two Cl- ions for each molecule, there will be 0.50 M concentration of both Mg2+ ions and twice as much, or 1.0 M, of Cl- ions in the new solution.

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

A sample of a hydrate of \(\mathrm{CuSO}_{4}\) with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? (A) \(\mathrm{CuSO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{CuSO}_{4} \cdot \mathrm{TH}_{2} \mathrm{O}\) (C) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen fluride with an enthalpy change of \(-540 \mathrm{kJ}\) . What is the value of the heat of formation of \(\mathrm{HF}(g) ?\) (A) \(-1,080 \mathrm{kJ} / \mathrm{mol}\) (B) \(-270 \mathrm{kJ} / \mathrm{mol}\) (C) 270 \(\mathrm{kJ} / \mathrm{mol}\) (D) 540 \(\mathrm{kJ} / \mathrm{mol}\)

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

A 1 -molar solution of a very weak monoprotic acid has a pH of 5. What is the value of \(K_{a}\) for the acid? (A) \(K_{a}=1 \times 10^{-10}\) (B) \(K_{a}=1 \times 10^{-7}\) (C) \(K_{a}=1 \times 10^{-5}\) (D) \(K_{a}=1 \times 10^{-2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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