Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 4, Problem 4.150P

(a)

Interpretation Introduction

Interpretation:

The balanced molecular, total ionic, and net ionic equations for the reaction that occurs when the solutions are mixed are to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

(a)

Expert Solution
Check Mark

Answer to Problem 4.150P

The balanced molecular equation of the reaction is as follows:

Na2CO3(aq)+CaCl2(aq)CaCO3(s)+2NaCl(aq)

The total ionic equation for the given reaction is as follows:

2Na+(aq)+CO32(aq)+Ca2+(aq)+2Cl(aq)CaCO3(s)+2Na+(aq)+2Cl(aq)

The net ionic equation for the given reaction is as follows:

CO32(aq)+Ca2+(aq)CaCO3(s)

Explanation of Solution

The first beaker consists of sodium carbonate (Na2CO3) and the second beaker consists of calcium chloride (CaCl2). The balanced molecular equation of the reaction is as follows:

Na2CO3(aq)+CaCl2(aq)CaCO3(s)+2NaCl(aq)

The total ionic equation for the given reaction is as follows:

2Na+(aq)+CO32(aq)+Ca2+(aq)+2Cl(aq)CaCO3(s)+2Na+(aq)+2Cl(aq)

Na+ and Cl are the spectator ions that are present in the reaction mixture. Spectator ions are not present in the net ionic equation.

The net ionic equation for the given reaction is as follows:

CO32(aq)+Ca2+(aq)CaCO3(s)

Conclusion

CaCO3 is the solid insoluble substance that is formed in the reaction. Na+ and Cl are the spectator ions that are present in the reaction mixture. These ions are present in the total ionic equation but are absent in the net ionic equation.

(b)

Interpretation Introduction

Interpretation:

Mass of precipitate formed when the yield is 100% is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The insoluble product is formed when the electrostatic attraction between the ions is greater as compared to the attraction between ions and water molecule. The product will remain intact and precipitate out from the solution.

(b)

Expert Solution
Check Mark

Answer to Problem 4.150P

Mass of precipitate formed when the yield is 100% is 10g.

Explanation of Solution

CaCO3 dissociates to give one Ca2+ and one CO32 ion. There are two Ca2+ ions and three CO32 ions in the solution so Ca2+ is the limiting reagent.

The formula to calculate the mass of CaCO3 is as follows:

Mass of CaCO3=[(numberof Ca2+sphere)(mole of1Ca2+sphere1sphere)(1molCaCO31molCa2+)(molar mass of CaCO3)] (1)

Substitute 2sphere for a number of Ca2+ sphere, 0.050mol for a mole of 1Ca2+ sphere and 100.09g for molar mass of CaCO3 in the equation (1).

Mass of CaCO3=[(2sphere)(0.050mol1sphere)(1molCaCO31molCa2+)(100.09g)]=10.009g10g

Conclusion

Mass of precipitate formed when the yield is 100% is 10g.

(c)

Interpretation Introduction

Interpretation:

The concentration of each ion in solution after reaction is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The insoluble product is formed when the electrostatic attraction between the ions is greater as compared to the attraction between ions and water molecule. The product will remain intact and precipitate out from the solution.

(c)

Expert Solution
Check Mark

Answer to Problem 4.150P

The concentration of Na+ is 0.60M, Cl is 0.40M, CO32 is 0.10M and Ca2+ is 0 after the reaction.

Explanation of Solution

The formula to calculate moles of Na+ is as follows:

Moles of Na+=[(numberof Na+sphere)(mole of1Na+sphere1sphere)] (2)

Substitute 6 spheres for number of Na+ sphere and 0.050mol for mole of 1Na+ sphere in the equation (2).

Moles of Na+=[(6 sphere)(0.050mol1sphere)]=0.30mol

The formula to calculate moles of CO32 is as follows:

Moles of CO32=[(numberof CO32sphere)(mole of1CO32sphere1sphere)] (3)

Substitute 3 spheres for number of CO32 sphere and 0.050mol for mole of 1CO32 sphere in the equation (3).

Moles of CO32=[(3 sphere)(0.050mol1sphere)]=0.15mol

The formula to calculate moles of Ca2+ is as follows:

Moles of Ca2+=[(numberof Ca2+sphere)(mole of1Ca2+sphere1sphere)] (4)

Substitute 2 spheres for number of Ca2+ sphere and 0.050mol for mole of 1Ca2+ sphere in the equation (4).

Moles of Ca2+=[(2 sphere)(0.050mol1sphere)]=0.10mol

The formula to calculate moles of Cl is as follows:

Moles of Cl=[(numberof Clsphere)(mole of1Clsphere1sphere)] (5)

Substitute 4 spheres for a number of Cl sphere and 0.050mol for mole of 1Cl sphere in the equation (5).

Moles of Cl=[(4 sphere)(0.050mol1sphere)]=0.20mol

After the reaction, the moles of Na+ and Cl remain the same. The moles of Ca2+ becomes zero as it is a limiting reagent. 0.10mol of Ca2+ reacts with 0.10mol of CO32.

The formula to calculate moles of CO32 left is as follows:

Moles ofCO32 left=Total moles ofCO32moles of CO32reacted (6)

Substitute 0.15mol for total moles of CO32 and 0.10mol for mole of CO32 reacted in the equation (6).

Moles ofCO32 left=0.15mol0.10mol=0.050mol

The total volume of solution is the sum of the volume of both the beaker that is 500mL.

The formula to calculate the molarity of Na+ is as follows:

Molarity of Na+=(moles of Na+volume of solution) (7)

Substitute 0.30mol for moles of Na+ and 500mL for the volume of solution in the equation (7).

Molarity of Na+=(0.30mol500mL)(1000mL1L)=0.60M

The formula to calculate the molarity of Cl is as follows:

Molarity of Cl=(moles of Clvolume of solution) (8)

Substitute 0.20mol for moles of Cl and 500mL for volume of solution in the equation (8).

Molarity of Cl=(0.20mol500mL)(1000mL1L)=0.40M

The formula to calculate molarity of CO32 is as follows:

Molarity of CO32=(moles of CO32volume of solution) (9)

Substitute 0.050mol for moles of CO32 and 500mL for volume of solution in the equation (9).

Molarity of CO32=(0.050mol500mL)(1000mL1L)=0.10M

Conclusion

The concentration of Na+ is 0.60M, Cl is 0.40M, CO32 is 0.10M and Ca2+ is 0 after the reaction.

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Chapter 4 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 4.1 - A chemist dilutes 60.0 mL of 4.50 M potassium...Ch. 4.1 - Prob. 4.6BFPCh. 4.1 - Prob. 4.7AFPCh. 4.1 - Prob. 4.7BFPCh. 4.3 - Prob. 4.8AFPCh. 4.3 - Prob. 4.8BFPCh. 4.3 - Prob. 4.9AFPCh. 4.3 - Molecular views of the reactant solutions for a...Ch. 4.3 - It is desirable to remove calcium ion from hard...Ch. 4.3 - To lift fingerprints from a crime scene, a...Ch. 4.3 - Despite the toxicity of lead, many of its...Ch. 4.3 - Mercury and its compounds have uses from fillings...Ch. 4.4 - How many OH−(aq) ions are present in 451 mL of...Ch. 4.4 - Prob. 4.12BFPCh. 4.4 - Prob. 4.13AFPCh. 4.4 - Prob. 4.13BFPCh. 4.4 - Prob. 4.14AFPCh. 4.4 - Prob. 4.14BFPCh. 4.4 - Another active ingredient in some antacids is...Ch. 4.4 - Prob. 4.15BFPCh. 4.4 - What volume of 0.1292 M Ba(OH)2 would neutralize...Ch. 4.4 - Calculate the molarity of a solution of KOH if...Ch. 4.5 - Prob. 4.17AFPCh. 4.5 - Prob. 4.17BFPCh. 4.5 - Prob. 4.18AFPCh. 4.5 - Prob. 4.18BFPCh. 4.5 - Prob. 4.19AFPCh. 4.5 - Prob. 4.19BFPCh. 4.6 - Prob. 4.20AFPCh. 4.6 - Prob. 4.20BFPCh. 4 - Prob. 4.1PCh. 4 - What types of substances are most likely to be...Ch. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Which of the following scenes best represents how...Ch. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - A mathematical equation useful for dilution...Ch. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Does an aqueous solution of each of the following...Ch. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Calculate each of the following quantities: Mass...Ch. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Calculate each of the following quantities: Volume...Ch. 4 - Prob. 4.30PCh. 4 - Concentrated sulfuric acid (18.3 M) has a density...Ch. 4 - Prob. 4.32PCh. 4 - Muriatic acid, an industrial grade of concentrated...Ch. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Write two sets of equations (both molecular and...Ch. 4 - Why do some pairs of ions precipitate and others...Ch. 4 - Use Table 4.1 to determine which of the following...Ch. 4 - The beakers represent the aqueous reaction of...Ch. 4 - Complete the following precipitation reactions...Ch. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - If 25.0 mL of silver nitrate solution reacts with...Ch. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - With ions shown as spheres and solvent molecules...Ch. 4 - The precipitation reaction between 25.0 mL of a...Ch. 4 - A 1.50-g sample of an unknown alkali-metal...Ch. 4 - Prob. 4.54PCh. 4 - The mass percent of Cl− in a seawater sample is...Ch. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Write a general equation for a neutralization...Ch. 4 - Prob. 4.59PCh. 4 - (a) Name three common weak acids. 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