Chemistry (AP Edition)
Chemistry (AP Edition)
9th Edition
ISBN: 9781133611103
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Brooks Cole
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Chapter 4, Problem 145MP

You have two 500.0-mL aqueous solutions. Solution A is a solution of a metal nitrate that is 8.246% nitrogen by mass. The ionic compound in solution B consists of potassium, chromium, and oxygen; chromium has an oxidation state of + 6 and there are 2 potassiums and 1 chromium in the formula. The masses of the solutes in each of the solutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction bas gone to completion, you dry the solid and find that it has a mass of 331.8 g.

a. Identify the ionic compounds in solution A and solution B.

b. Identify the blood-red precipitate.

c. Calculate the concentration (molarity) of all ions in the original solutions.

d. Calculate the concentration (molarity) of all ions in the final solution.

a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The ionic compounds in solution A and B has to be identified.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

Concentration(M)=Molesofsolute(g)Volumeofsolution(L)

Answer to Problem 145MP

The ionic compound in solution A is AgNO3 and ionic compound B is K2CrO4 .

Explanation of Solution

Given:

Record the given data

Volume of two aqueous solution   = 500.0mL

Mass percent of Nitrogen in solution A = 8.246%

Mass of precipitate   = 331.8g

The volume of two aqueous solutions along with mass percent of Nitrogen in solution A of metal nitrate and mass of precipitate are recorded as shown above.

To identify the ionic compound in solution A

Molecular mass of oxygen = 16.00g

Molecular mass of nitrogen = 14.01g

Compound A contains metal nitrate M(NO3)x

Assuming the mass of metal nitrate is M(NO3)x = 100.0g

The moles of oxygen is calculated using

Molesofoxygen=8.246g48.00gO14.01gN=28.25gO

Thus, the mass of nitrate = 8.246+28.25g=36.50g(ifx=1)

Case 1:

x=1:massofM=100.00-36.50g=63.50g

MolM=molN=8.246g14.01g/mol=0.5886mol

MolarmassofmetalM=63.50g0.5886mol=107.9g/mol

The metal M is identified as Silver (Ag)

Case 2:

x=2:MassofM=100.00-2(36.50)=27.00g

MolM=12molN=0.5886mol2=0.2943mol

MolarmassofmetalM=27.00g0.2943mol=91.74g/mol

This is close to Zr but Zr doesn’t form stable +2 ions in the solution, and therefore it forms stable +4 ions.

x would not take the value of 3, since the nitrates present would have a higher mass than 100.0g .

Therefore, the compound A is identified as AgNO3

The ionic compound is solution A is calculated by plugging in the values of mass of Nitrogen in metal nitrate and also by calculating the mass of M and its molar mass. The value of calculated molar mass is matched with value of closest molar mass of element in the periodic table. The ionic compound is identified as AgNO3 .

To identify the compound B

Oxidation state of chromium = +6

The compound B has two Potassium atoms and one Chromium atom. Hence, the formula can be given as, K2CrOx .

The oxidation state of oxygen is calculated by,

6+x(-2)=-2x=4

Therefore, the formula becomes K2CrO4

Compound B is identified K2CrO4

The ionic compound in solution B is identified by calculating the oxidation state of Oxygen and substituting in the unknown x in the formula. The ionic compound in solution B is identified as K2CrO4 .

b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The blood-red precipitate has to be identified.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

Concentration(M)=Molesofsolute(g)Volumeofsolution(L)

Answer to Problem 145MP

The blood red precipitate is Ag2CrO4 .

Explanation of Solution

To identify the blood red precipitate

The blood red precipitate is identified as Silver chromate. A precipitate of silver chromate is formed when aqueous solution of Silver nitrate is reacted with aqueous solution of Potassium chromate. Potassium nitrate remains as spectator ions inside the solution.

The chemical equation can be given as,

2AgNO3(aq)+K2CrO4(aq)Ag2CrO4(s)+2KNO3(aq)

c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The concentration of original solution has to be calculated.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

Concentration(M)=Molesofsolute(g)Volumeofsolution(L)

Answer to Problem 145MP

The concentration of all ions in original solution are

SolutionA,Ag+=4.000MAg+NO3-=4.000MNO3-SolutionB,K+=7.000MK+CrO42-=3.500MCrO42-

Explanation of Solution

Calculate the concentration of ions in original solutions.

Mass of precipitate = 331.8g

Molar mass of AgNO3=169.9g

Molar mass of K2CrO4=194.2g

The chemical equation can be given as,

2AgNO3(aq)+K2CrO4(aq)Ag2CrO4(s)+2KNO3(aq)

From the chemical equation, 2 moles of AgNO3 and 1 mole of K2CrO4 reacts to yield 1 mole ( 331.8g ) of Ag2CrO4(s) .1 mole of precipitate is formed since 331.8g of Ag2CrO4(s) is produced that is equal to molar mass of Ag2CrO4(s) .

MassofAgNO3=2.000molAgNO3×169.9gmol=339.8gAgNO3Mass ofK2CrO4=1.000molK2CrO4×194.2gmol=194.2gK2CrO4

From the given question, the reactants are said to have equal masses.

Therefore, the sum of reactant masses must be either,

339.8gAgNO3+339.8gK2CrO4 or  194.2gAgNO3+194.2gK2CrO4

Assuming that, 194.2g quantities are correct

Then, when 194.2gK2CrO4(1mol)reactswith339.8gAgNO3(2mol) must be present to react with all K2CrO4 .

But we assumed only 194.2 g AgNO3 is present; this cannot be correct. Instead of K2CrO4 limiting, AgNO3 must be limiting, and we reacted 339.8 g AgNO3 and 339.8 g K2CrO4

Therefore, the concentration of ions in original solution,

SolutionA,Molarity of Ag+=2.000molAg+0.5000L=4.000MAg+MolarityofNO3-=2.000molNO3-0.5000L=4.000MNO3-SolutionB,Moles of K2CrO4=339.8gK2CrO4×1mol194.2g=1.750molK2CrO4MolarityofK+=2×1.750molK+0.5000L=7.000MK+MolarityofCrO42-=1.750molCrO42-0.5000L=3.500MCrO42-

The concentrations of individual ions are calculated by plugging in the values of moles taking part in the chemical reaction to the volumes of solutions. The concentrations of individual ions in original solution are found to be 4.000MAg+ , 4.000MNO3- , 7.000MK+ and 3.500MCrO42-

d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The concentration of final solution has to be calculated.

Concept Introduction: When two soluble solutions are mixed together, an insoluble salt formation occurs called as precipitate. These precipitates fall out of the solution and the reactions are called as precipitation reaction.

Concentration of solution can be defined in terms of molarity as moles of solute (in grams) to the volume of solution (in litres). The concentration of solution can be calculated by,

Concentration(M)=Molesofsolute(g)Volumeofsolution(L)

Answer to Problem 145MP

The concentration of all ions in final solution are

                         MK+=3.500MK+MNO3-=2.000MNO3-MCrO42-=0.750MCrO42-MAg+=0M(limitingreagent)

Explanation of Solution

To calculate the concentration of ions in final solution

The moles of spectator ions ( K+andNO3- ) remains unchanged after the reaction.

Ag+ is limiting reagent and hence the molarity of Ag+ will be zero after the precipitation is complete.

Ag+(aq)+CrO42-(aq)Ag2CrO4(s)Initial2.000mol1.750mol0Change-2.000mol-1.000mol+1.000molAfterreaction00.750mol1.000mol

Therefore, the concentrations of individual ions in final solution are

MolarityofK+=2×1.750mol1.0000L=3.500MK+MolarityofNO3-=2.000mol1.0000L=2.000MNO3-MolarityofCrO42-=0.750mol1.0000L=0.750MCrO42-MolarityofAg+=0M(limitingreagent)

The concentrations of individual ions in final solution are calculated by plugging in values of their respective moles to the volume of the solution. The molarity of individual ions in final solution are found to be 3.500MK+ , 2.000MNO3- , 0.750MCrO42- and 0M Ag+ respectively.

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Chapter 4 Solutions

Chemistry (AP Edition)

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