/*! 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 70 In each of the following pairs, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 70

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{I}^{-}\) ion: (a) 0.10 \(\mathrm{M}\) BaI \(_{2}\) or 0.25 \(\mathrm{M}\) KI solution, (b) 100 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) KI solution or 200 \(\mathrm{mL}\) of 0.040 \(\mathrm{MZnI}_{2}\) solution, \((\mathbf{c}) 3.2 \mathrm{M}\) HI solution or a solution made by dissolving 145 g of Nal in water to make 150 \(\mathrm{mL}\) of solution.

Short Answer

Expert verified
(a) 0.25 M KI solution has a higher concentration of I鈦� (0.25 M) compared to 0.10 M BaI鈧� solution (0.20 M). (b) 200 mL of 0.040 M ZnI鈧� solution has a higher concentration of I鈦� (0.016 mol) compared to 100 mL of 0.10 M KI solution (0.01 mol). (c) 6.45 M NaI solution has a higher concentration of I鈦� compared to the 3.2 M HI solution.

Step by step solution

01

(a) Comparing 0.10 M BaI2 and 0.25 M KI solutions

BaI2 (barium iodide) dissociates into Ba虏鈦� and 2I鈦� ions in the solution. Therefore, for every 1 mole of BaI2 dissolved, two moles of I鈦� are released. On the other hand, KI (potassium iodide) dissociates into K鈦� and I鈦� ions in the solution. Hence, for every 1 mole of KI dissolved, one mole of I鈦� is released. For BaI2: \( 0.10\, M \times 2 = 0.20\, M\) For KI: \( 0.25\, M \times 1 = 0.25\, M\) Thus, 0.25 M KI solution has the higher concentration of I鈦�.
02

(b) Comparing 100 mL of 0.10 M KI and 200 mL of 0.040 M ZnI鈧� solutions

As mentioned earlier, 1 mole of KI releases 1 mole of I鈦�, while 1 mole of ZnI鈧� (zinc iodide) releases 2 moles of I鈦� upon dissociation. For KI, iodide concentration in 100 mL solution (0.10 L) is: \(0.10\,M \times 0.10 L = 0.01 mol\) For ZnI鈧�, iodide concentration in 200 mL solution (0.20 L) is: \(2 \times 0.040 M \times 0.20 L = 0.016 mol\) Since 0.016 mol of I鈦� in 200 mL ZnI鈧� > 0.01 mol of I鈦� in 100 mL KI, 200 mL of 0.040 M ZnI鈧� solution has a higher concentration of I鈦�.
03

(c) Comparing 3.2 M HI and 145 g NaI in 150 mL solutions

1 mole of HI (hydroiodic acid) dissociates into 1 mole of I鈦� upon dissociation. Molar concentration of I鈦� in HI: \(3.2\,M HI \times 1 = 3.2\,M\ I^{-}\) For NaI (sodium iodide), we first find the number of moles in 145 g of NaI. Molar mass of NaI = 22.99 (Na) + 126.90 (I) = 149.89 g/mol Number of moles of NaI: \(\frac{145 g}{149.89 \frac{g}{mol}} = 0.967\, mol\) For every 1 mole of NaI dissolved, 1 mole of I鈦� is released. Molar concentration of I鈦� in NaI: \(\frac{0.967\, mol}{0.150 L} = 6.45\,M\) Thus, the 6.45 M NaI solution has a higher concentration of I鈦� compared to the 3.2 M HI 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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Dissociation of Ionic Compounds
Understanding how ionic compounds dissociate in water is crucial for grasping the concentration of ions in a solution. When an ionic compound such as KI dissolves in water, it separates into its constituent ions, in this case, potassium (K鈦�) and iodide (I鈦�). The ratio of these ions depends on the compound's chemical formula. For example, BaI2 splits into one barium ion (Ba虏鈦�) and two iodide ions (2I鈦�), releasing twice the number of iodide ions into the solution per formula unit compared to KI, which only releases one iodide ion.

Dissociation is a reversible process where the solid form and dissolved ions are in dynamic equilibrium in a saturated solution. However, this removal of ions from the equilibrium by reactions or dilutions drives the reaction toward complete dissociation, making it essentially irreversible in dilute solutions. This knowledge helps to anticipate the concentration of specific ions in a solution, a vital concept in understanding chemical reactions in a solution.
Molar Concentration
Molar concentration鈥攁lso known as molarity鈥攊s the number of moles of a solute per liter of solution. It is expressed in units of moles per liter (M). Calculating the molar concentration is straightforward when you have a pure compound: simply divide the amount of substance in moles by the volume of the solution in liters. For ionic compounds that dissociate, each ion type must be considered separately. When determining which solution has a higher concentration of a specific ion, you need to factor in not only the molarity of the ionic compound but also the dissociation pattern.

For example, as highlighted in the given solution for BaI2, you multiply the molarity of the compound by the number of the specific ion it releases upon dissociation. The product yields the concentration of that ion in the solution. Recognizing these relationships helps in comparing different solutions to determine the one with a higher concentration of a particular ion.
Stoichiometry of Dissolution
The stoichiometry of dissolution is a part of chemical stoichiometry that focuses on the proportions in which substances dissolve to form solutions. The stoichiometry of an ionic compound provides the ratio of cations to anions released upon dissolution. For example, with ZnI2, the ratio is 1:2 (one Zn虏鈦� to two I鈦� ions). This stoichiometry is essential for calculating the concentration of ions after dissolution.

In practical terms, if you dissolve 1 mole of ZnI2, you obtain 2 moles of iodide ions. So, for a solution with a known volume, you multiply the molarity of ZnI2 by 2 to find the molarity of I鈦� in that solution. Similarly, the calculation for the molar concentration of I鈦� from NaI involves using the molar mass of NaI and the weight of NaI to find the moles of solute, which then allows for the calculation of molarity when you divide by the solution volume. This highlights why understanding stoichiometry is pivotal for determining concentrations in chemistry.

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

The following reactions (note that the arrows are pointing only one direction) can be used to prepare an activity series for the halogens: $$\begin{array}{c}{\mathrm{Br}_{2}(a q)+2 \mathrm{Nal}(a q) \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q)} \\ {\mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q)}\end{array}$$ (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.

Write balanced molecular and net ionic equations for the following reactions, and identify the gas formed in each: (a) solid cadmium sulfide reacts with an aqueous solution of sulfuric acid; (b) solid magnesium carbonate reacts with an aqueous solution of perchloric acid.

Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(\mathrm{s})\) ; solutions of metals like this that are solids are called alloys. There is a range of compositions over which the solution is considered a bronze. Bronzes are stronger and harder than either copper or tin alone. (a) \(\mathrm{A} 100.0\) -g sample of a certain bronze is 90.0\(\%\) copper by mass and 10.0\(\%\) tin. Which metal can be called the solvent, and which the solute? (b) Based on part (a), calculate the concentration of the solute metal in the alloy in units of molarity, assuming a density of 7.9 \(\mathrm{g} / \mathrm{cm}^{3}\) . (c) Suggest a reaction that you could do to remove all the tin from this bronze to leave a pure copper sample. Justify your reasoning.

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

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.