Concept explainers
(a)
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)
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)+CO2−3(aq)+Ca2+(aq)+2Cl−(aq)→CaCO3(s)+2Na+(aq)+2Cl−(aq)
The net ionic equation for the given reaction is as follows:
CO2−3(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)+CO2−3(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:
CO2−3(aq)+Ca2+(aq)→CaCO3(s)
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:
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)
Answer to Problem 4.150P
Mass of precipitate formed when the yield is 100 % is 10 g.
Explanation of Solution
CaCO3 dissociates to give one Ca2+ and one CO2−3 ion. There are two Ca2+ ions and three CO2−3 ions in the solution so Ca2+ is the limiting reagent.
The formula to calculate the mass of CaCO3 is as follows:
Mass of CaCO3=[(number of Ca2+ sphere)(mole of 1 Ca2+ sphere1 sphere)(1 mol CaCO31 mol Ca2+ )(molar mass of CaCO3)] (1)
Substitute 2 sphere for a number of Ca2+ sphere, 0.050 mol for a mole of 1Ca2+ sphere and 100.09 g for molar mass of CaCO3 in the equation (1).
Mass of CaCO3=[(2 sphere)(0.050 mol1 sphere)(1 mol CaCO31 mol Ca2+ )(100.09 g)]=10.009 g≈10 g
Mass of precipitate formed when the yield is 100 % is 10 g.
(c)
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)
Answer to Problem 4.150P
The concentration of Na+ is 0.60 M, Cl− is 0.40 M, CO2−3 is 0.10 M and Ca2+ is 0 after the reaction.
Explanation of Solution
The formula to calculate moles of Na+ is as follows:
Moles of Na+=[(number of Na+ sphere)(mole of 1 Na+ sphere1 sphere)] (2)
Substitute 6 spheres for number of Na+ sphere and 0.050 mol for mole of 1Na+ sphere in the equation (2).
Moles of Na+=[(6 sphere)(0.050 mol1 sphere)]=0.30 mol
The formula to calculate moles of CO2−3 is as follows:
Moles of CO2−3=[(number of CO2−3 sphere)(mole of 1 CO2−3 sphere1 sphere)] (3)
Substitute 3 spheres for number of CO2−3 sphere and 0.050 mol for mole of 1CO2−3 sphere in the equation (3).
Moles of CO2−3=[(3 sphere)(0.050 mol1 sphere)]=0.15 mol
The formula to calculate moles of Ca2+ is as follows:
Moles of Ca2+=[(number of Ca2+ sphere)(mole of 1 Ca2+ sphere1 sphere)] (4)
Substitute 2 spheres for number of Ca2+ sphere and 0.050 mol for mole of 1Ca2+ sphere in the equation (4).
Moles of Ca2+=[(2 sphere)(0.050 mol1 sphere)]=0.10 mol
The formula to calculate moles of Cl− is as follows:
Moles of Cl−=[(number of Cl− sphere)(mole of 1 Cl− sphere1 sphere)] (5)
Substitute 4 spheres for a number of Cl− sphere and 0.050 mol for mole of 1Cl− sphere in the equation (5).
Moles of Cl−=[(4 sphere)(0.050 mol1 sphere)]=0.20 mol
After the reaction, the moles of Na+ and Cl− remain the same. The moles of Ca2+ becomes zero as it is a limiting reagent. 0.10 mol of Ca2+ reacts with 0.10 mol of CO2−3.
The formula to calculate moles of CO2−3 left is as follows:
Moles of CO2−3 left=Total moles of CO2−3− moles of CO2−3 reacted (6)
Substitute 0.15 mol for total moles of CO2−3 and 0.10 mol for mole of CO2−3 reacted in the equation (6).
Moles of CO2−3 left=0.15 mol− 0.10 mol=0.050 mol
The total volume of solution is the sum of the volume of both the beaker that is 500 mL.
The formula to calculate the molarity of Na+ is as follows:
Molarity of Na+=(moles of Na+volume of solution) (7)
Substitute 0.30 mol for moles of Na+ and 500 mL for the volume of solution in the equation (7).
Molarity of Na+=(0.30 mol500 mL)(1000 mL1 L)=0.60 M
The formula to calculate the molarity of Cl− is as follows:
Molarity of Cl−=(moles of Cl−volume of solution) (8)
Substitute 0.20 mol for moles of Cl− and 500 mL for volume of solution in the equation (8).
Molarity of Cl−=(0.20 mol500 mL)(1000 mL1 L)=0.40 M
The formula to calculate molarity of CO2−3 is as follows:
Molarity of CO2−3=(moles of CO2−3volume of solution) (9)
Substitute 0.050 mol for moles of CO2−3 and 500 mL for volume of solution in the equation (9).
Molarity of CO2−3=(0.050 mol500 mL)(1000 mL1 L)=0.10 M
The concentration of Na+ is 0.60 M, Cl− is 0.40 M, CO2−3 is 0.10 M and Ca2+ is 0 after the reaction.
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Chapter 4 Solutions
Student Solutions Manual For Silberberg Chemistry: The Molecular Nature Of Matter And Change With Advanced Topics
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