Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
Question
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Chapter 6, Problem 32E

a)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

Molarity(inM)=massofsolute(in g)volumeofsolution(inL)

a)

Expert Solution
Check Mark

Answer to Problem 32E

0.0200 mole of Sodium phosphate in 10.0 mL of solution.

MNa+6.00MMPO43-2.00M

Explanation of Solution

Record the given data

Moles of Sodium phosphate= 0.200moles

Volume of the solution= 10.0mL

Calculation for the concentration of separate ions is as follows,

The balanced equation for dissolving ions,

Na3PO4(s)3Na+(aq)+PO43-(aq)

The molarity of ions can be calculated by the formula,

Molarity=massofsolutevolumeofsolution

Therefore, by substituting the given info in the formula, the concentrations of separate ions can be given as,

Molarity=0.02000.0100=2.00M

In order, to find the concentration of the separate ion,

Concentrationofindividualion=numberofatoms×thecalculatedmolarity

Therefore, the concentrations of separate ions can be given by,

3Na+=3×2.00=6.00MNa+PO43-=1×3.00=2.00MPO43-

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual Na+ and PO43- are found to be 6.00MNa+ and 2.00MPO43- respectively.

b)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

Molarity(inM)=massofsolute(in g)volumeofsolution(inL)

b)

Expert Solution
Check Mark

Answer to Problem 32E

0.300 mole of Barium nitrate in 600 mL of solution.

MBa2+0.500MMNO3-1.00M

Explanation of Solution

Record the given data

Moles of Barium nitrate= 0.300mole

Volume of the solution= 600mL

Calculation for the concentration of separate ions is as follows,

The balanced equation for dissolving ions

Ba(NO3)2(s)Ba2++2NO32-

The molarity of ions can be calculated by the formula,

Molarity=massofsolutevolumeofsolution

Therefore, by substituting the given info in the formula, molarity can be calculated by,

Molarity=0.3000.600=0.500M

In order, to find the concentration of the separate ion,

Concentrationofindividualion=numberofatoms×thecalculatedmolarity

Therefore, the concentrations of separate ions can be given by,

Ba2+=1×0.5=0.5MBa2+NO3-=2×0.5=1.00MNO3-

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual Ba2+ and NO3- are 0.5MBa2+ and 1.00MNO3- respectively.

c)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

Molarity(inM)=massofsolute(in g)volumeofsolution(inL)

c)

Expert Solution
Check Mark

Answer to Problem 32E

1.00 g of Potassium chloride in 0.500 L of solution.

MK+0.0268MMCl-0.0268M

Explanation of Solution

Calculation

Record the given info

Mole of Potassium chloride=1.00 gram

Volume of solution= 0.500 L

for the concentration of individual ions is as follows,

The balanced equation can be given as,

KClK++Cl-

In order to calculate molarity, grams are converted into moles by using the molar mass.

The molar mass can be calculated by sum of mass of individual elements.

Molar mass of Potassium chloride can be given as (1×40) + (1×35.5) =74.55 g/mol

Amount of Potassium chloride= MassMolarMass

                                                  = 174.55=0.01341mol

Concentration=amountvolume =0.013410.500=0.0268M

In order, to find the concentration of the separate ion,

Concentrationofindividualion=numberofatoms×thecalculatedmolarity

Therefore, the concentrations of separate ions can be given by,

K+=1×0.0268=0.0268MK+Cl-=1×0.0268=0.0268MCl-

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual K+ and Cl- are 0.0268MK+ and 0.0268MCl-

d)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

Molarity(inM)=massofsolute(in g)volumeofsolution(inL)

d)

Expert Solution
Check Mark

Answer to Problem 32E

132 g of Ammonium sulphate in 1.50 L of solution.

MNH4+0.66MMSO42-1.32M

Explanation of Solution

Record the given info

Mole of Ammonium sulphate=132 gram

Volume of solution=1.50L

Calculation for the concentration of individual ions is as follows,

The balanced equation for dissolving ions,

(NH4)2SO42NH42++SO42-

In order to calculate molarity, grams are converted into moles by using the molar mass.

The molar mass can be calculated by sum of mass of individual elements.

Molar mass of Ammonium sulphate is 132 g/mol

AmountofAmmoniumsulphate = MassMolarMass

AmountofAmmoniumsulphate=132132=1.00mol

Therefore, from the amount of Ammonium sulphate, the molarity can be calculated as,

Concentration=amountvolume

Concentration=11.5=0.666M

In order, to find the concentration of the separate ion,

Concentrationofindividualion=numberofatoms×thecalculatedmolarity

Therefore, the concentrations of separate ions can be given by,

NH4+=2×0.666=1.32MNH4+SO42-=1×0.666=0.666MSO42-

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual NH4+ and SO42- are

1.32MNH4+ and 0.666MSO42- .

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Chapter 6 Solutions

Chemistry: An Atoms First Approach

Ch. 6 - Prob. 2ALQCh. 6 - You have a sugar solution (solution A) with...Ch. 6 - Prob. 4ALQCh. 6 - Prob. 5ALQCh. 6 - Prob. 6ALQCh. 6 - Consider separate aqueous solutions of HCl and...Ch. 6 - Prob. 8ALQCh. 6 - Prob. 9ALQCh. 6 - The exposed electrodes of a light bulb are placed...Ch. 6 - Differentiate between what happens when the...Ch. 6 - Consider the following electrostatic potential...Ch. 6 - Prob. 15QCh. 6 - A typical solution used in general chemistry...Ch. 6 - Prob. 17QCh. 6 - A student wants to prepare 1.00 L of a 1.00-M...Ch. 6 - List the formulas of three soluble bromide salts...Ch. 6 - When 1.0 mole of solid lead nitrate is added to...Ch. 6 - What is an acid and what is a base? An acid-base...Ch. 6 - A student had 1.00 L of a 1.00-M acid solution....Ch. 6 - Prob. 23QCh. 6 - Prob. 24QCh. 6 - Prob. 25ECh. 6 - Match each name below with the following...Ch. 6 - Prob. 27ECh. 6 - Commercial cold packs and hot packs are available...Ch. 6 - Calculate the molarity of each of these solutions....Ch. 6 - A solution of ethanol (C2H5OH) in water is...Ch. 6 - Calculate the concentration of all ions present in...Ch. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Prob. 34ECh. 6 - Prob. 35ECh. 6 - Prob. 36ECh. 6 - Prob. 37ECh. 6 - Prob. 38ECh. 6 - A solution is prepared by dissolving 10.8 g...Ch. 6 - A solution was prepared by mixing 50.00 mL of...Ch. 6 - Calculate the sodium ion concentration when 70.0...Ch. 6 - Suppose 50.0 mL of 0.250 M CoCl2 solution is added...Ch. 6 - Prob. 43ECh. 6 - A stock solution containing Mn2+ ions was prepaned...Ch. 6 - On the basis of the general solubility rules given...Ch. 6 - On the basis of the general solubility rules given...Ch. 6 - When the following solutions are mixed together,...Ch. 6 - When the following solutions are mixed together,...Ch. 6 - For the reactions in Exercise 47, write the...Ch. 6 - For the reactions in Exercise 48, write the...Ch. 6 - Write the balanced formula and net ionic equation...Ch. 6 - Give an example how each of the following...Ch. 6 - Write net ionic equations for the reaction, if...Ch. 6 - Write net ionic equations for the reaction, if...Ch. 6 - Prob. 55ECh. 6 - Prob. 56ECh. 6 - What mass of Na2CrO4 is required to precipitate...Ch. 6 - What volume of 0.100 M Na3PO4 is required to...Ch. 6 - What mass of solid aluminum hydroxide can be...Ch. 6 - What mass of barium sulfate can be produced when...Ch. 6 - What mass of solid AgBr is produced when 100.0 mL...Ch. 6 - What mass of silver chloride can be prepared by...Ch. 6 - A 100.0-mL aliquot of 0.200 M aqueous potassium...Ch. 6 - A 1.42-g sample of a pure compound, with formula...Ch. 6 - You are given a 1.50-g mixture of sodium nitrate...Ch. 6 - Write the balanced formula, complete ionic, and...Ch. 6 - Write the balanced formula, complete ionic, and...Ch. 6 - Write the balanced formula equation for the...Ch. 6 - Prob. 70ECh. 6 - What volume of each of the following acids will...Ch. 6 - Prob. 72ECh. 6 - Hydrochloric acid (75.0 mL of 0.250 M) is added to...Ch. 6 - Prob. 74ECh. 6 - A 25.00-mL sample of hydrochloric acid solution...Ch. 6 - A 10.00-mL sample of vinegar, an aqueous solution...Ch. 6 - What volume of 0.0200 M calcium hydroxide is...Ch. 6 - A 30.0-mL sample of an unknown strong base is...Ch. 6 - A student titrates an unknown amount of potassium...Ch. 6 - The concentration of a certain sodium hydroxide...Ch. 6 - Assign oxidation states for all atoms in each of...Ch. 6 - Assign the oxidation state for nitrogen in each of...Ch. 6 - Prob. 84ECh. 6 - Specify which of the following are...Ch. 6 - Specify which of the following equations represent...Ch. 6 - Consider the reaction between sodium metal and...Ch. 6 - Consider the reaction between oxygen (O2) gas and...Ch. 6 - Balance each of the following oxidationreduction...Ch. 6 - Balance each of the following oxidationreduction...Ch. 6 - Prob. 91AECh. 6 - Prob. 92AECh. 6 - Prob. 93AECh. 6 - Prob. 94AECh. 6 - Prob. 95AECh. 6 - Consider a 1.50-g mixture of magnesium nitrate and...Ch. 6 - A 1.00-g sample of an alkaline earth metal...Ch. 6 - A mixture contains only NaCl and Al2(SO4)3. A...Ch. 6 - The thallium (present as Tl2SO4) in a 9.486-g...Ch. 6 - Prob. 100AECh. 6 - A student added 50.0 mL of an NaOH solution to...Ch. 6 - Prob. 102AECh. 6 - Acetylsalicylic acid is the active ingredient in...Ch. 6 - When hydrochloric acid reacts with magnesium...Ch. 6 - A 2.20-g sample of an unknown acid (empirical...Ch. 6 - Carminic acid, a naturally occurring red pigment...Ch. 6 - Chlorisondamine chloride (C14H20Cl6N2) is a drug...Ch. 6 - Prob. 108AECh. 6 - Prob. 109AECh. 6 - Many oxidationreduction reactions can be balanced...Ch. 6 - Prob. 111AECh. 6 - Calculate the concentration of all ions present...Ch. 6 - A solution is prepared by dissolving 0.6706 g...Ch. 6 - For the following chemical reactions, determine...Ch. 6 - What volume of 0.100 M NaOH is required to...Ch. 6 - Prob. 116CWPCh. 6 - A 450.0-mL sample of a 0.257-M solution of silver...Ch. 6 - The zinc in a 1.343-g sample of a foot powder was...Ch. 6 - Prob. 119CWPCh. 6 - When organic compounds containing sulfur are...Ch. 6 - Prob. 121CWPCh. 6 - Prob. 122CPCh. 6 - The units of parts per million (ppm) and parts per...Ch. 6 - Prob. 124CPCh. 6 - Prob. 125CPCh. 6 - Prob. 126CPCh. 6 - Consider the reaction of 19.0 g of zinc with...Ch. 6 - A mixture contains only sodium chloride and...Ch. 6 - Prob. 129CPCh. 6 - Prob. 130CPCh. 6 - Prob. 131CPCh. 6 - Consider reacting copper(II) sulfate with iron....Ch. 6 - Prob. 133CPCh. 6 - Prob. 134CPCh. 6 - What volume of 0.0521 M Ba(OH)2 is required to...Ch. 6 - A 10.00-mL sample of sulfuric acid from an...Ch. 6 - Prob. 137CPCh. 6 - A 6.50-g sample of a diprotic acid requires 137.5...Ch. 6 - Citric acid, which can be obtained from lemon...Ch. 6 - Prob. 140CPCh. 6 - Prob. 141CPCh. 6 - Tris(pentatluorophenyl)borane, commonly known by...Ch. 6 - In a 1-L beaker, 203 mL of 0.307 M ammonium...Ch. 6 - The vanadium in a sample of ore is converted to...Ch. 6 - The unknown acid H2X can be neutralized completely...Ch. 6 - Three students were asked to find the identity of...Ch. 6 - You have two 500.0-mL aqueous solutions. Solution...
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