/*! 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 17 Assuming the volumes are additiv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 17

Assuming the volumes are additive, what is the \(\left[\mathrm{Cl}^{-}\right]\) in a solution obtained by mixing \(225 \mathrm{mL}\) of \(0.625 \mathrm{M}\) \(\mathrm{KCl}\) and \(615 \mathrm{mL}\) of \(0.385 \mathrm{M} \mathrm{MgCl}_{2} ?\)

Short Answer

Expert verified
The concentration of Chloride ions [Cl-] in the resulting solution would be approximately 0.7314 M.

Step by step solution

01

Understand The Chemical Compounds

The compound KCl dissociates in solution to form K+ and Cl- ions. On the other hand, MgCl2 gives us Mg2+ and two Cl- ions in solution. This means that a mole of KCl will have the same number of moles of Cl- ions while a mole of MgCl2 will have twice the number of moles of Cl- ions.
02

Calculate The Moles

For KCl, the number of moles can be calculated by multiplying the volume of the KCl solution (in liters) by its molar concentration. Thus, moles of Cl- from KCl = 225 mL * (1L/1000 mL) * 0.625 M = 0.140625 moles. We can obtain the moles of chloride ions from MgCl2 in the same way: moles of Cl- from MgCl2 = 615 mL * (1 L/1000 mL) * 0.385 M * 2 = 0.47355 moles. The multiplication by 2 accounts for the fact that each molecular unit of MgCl2 gives 2 ions of Cl-.
03

Add The Moles

Now we need to add the moles of Cl- ions from both solutions to get the total moles. Total moles of Cl- = moles of Cl- from KCl + moles of Cl- from MgCl2 = 0.140625 moles + 0.47355 moles = 0.614175 moles.
04

Calculate The Concentration

Finally, the molar concentration (M) of Cl- ions can be obtained by dividing the total moles by the total volume. The total volume = 225 mL + 615 mL = 840 mL = 0.84 L. Therefore, molar concentration (M) = total moles / total volume in Liters = 0.614175 moles / 0.84 L = 0.7314 M.

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.

Chloride Ion Concentration
Understanding chloride ion concentration in a solution involves recognizing how different compounds contribute their ions when dissolved. In this exercise, we mix solutions of KCl and MgCl鈧�. Both compounds dissolve and dissociate in water to release chloride ions (Cl鈦�).
  • KCl: Each molecule of potassium chloride (KCl) dissociates into one K鈦� ion and one Cl鈦� ion.
  • 惭驳颁濒鈧�: Magnesium chloride (MgCl鈧�) behaves differently. It splits into one Mg虏鈦� ion and two Cl鈦� ions. Thus, MgCl鈧� contributes twice the number of Cl鈦� ions per molecule compared to KCl.
Careful consideration of the dissociation process is key as it influences how many chloride ions are present in the final solution. Knowing the behavior of these molecules upon dissolving helps determine the total chloride ion concentration.
Molarity Calculation
Calculating molarity, which measures concentration, is essential when mixing solutions. Molarity is defined as the number of moles of solute per liter of solution. To find it, we need:
  • Total moles of solute (chloride ions in this case)
  • Volume of the solution in liters
For example, with KCl's molarity known and its volume converted to liters, we calculate the moles of Cl鈦� ions. The same process applies to MgCl鈧�, multiplying by 2 for its two chloride ions per formula unit. Finally, total moles of chloride ions are found by adding the moles from both solutions. Dividing this total by the solution's overall volume gives molarity. This calculation underscores how solute and solvent quantities relate to the concentration of the chloride ions.
Chemical Dissociation
Chemical dissociation refers to the process where ionic compounds separate into individual ions in solution. This understanding is vital for predicting the behavior of dissolved substances.
  • When KCl dissociates, it directly forms one Cl鈦� ion per molecule.
  • However, MgCl鈧�'s dissociation results in two Cl鈦� ions per molecule, doubling its contribution of chloride ions relative to KCl.
Knowing how compounds dissociate allows for accurate calculation of ion concentrations in solutions. This principle not only helps in chemistry exercises but also in real-world applications like determining pollutant concentrations in water supplies or medication dosages. By grasping the basics of dissociation, we understand the fundamental chemical behaviors necessary for calculating subsequent concentrations such as molarity.

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 these equations for redox reactions occurring in acidic solution. (a) \(\mathrm{MnO}_{4}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{I}_{2}(\mathrm{s})\) (b) \(\mathrm{BrO}_{3}^{-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{Br}^{-}+\mathrm{N}_{2}\) (c) \(\mathrm{VO}_{4}^{3-}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{VO}^{2+}+\mathrm{Fe}^{3+}\) (d) \(\mathrm{UO}^{2+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{UO}_{2}^{2+}+\mathrm{NO}(\mathrm{g})\)

Assign oxidation states to the elements involved in the following reactions. Indicate which are redox reactions and which are not. (a) \(\mathrm{MgCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})\) (c) \(\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}_{2}(\mathrm{g})\) (d) \(2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\)

A \(7.55 \mathrm{g}\) sample of \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is added to \(125 \mathrm{mL}\) of a vinegar that is \(0.762 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} .\) Will the resulting solution still be acidic? Explain.

An iron ore sample weighing \(0.9132 \mathrm{g}\) is dissolved in \(\mathrm{HCl}(\mathrm{aq}),\) and the iron is obtained as \(\mathrm{Fe}^{2+}(\mathrm{aq}) .\) This solution is then titrated with \(28.72 \mathrm{mL}\) of \(0.05051 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) What is the mass percent Fe in the ore sample? \(6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow_{6 \mathrm{Fe}^{3+}}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

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.