Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
Question
Chapter 15, Problem 14ALQ
Interpretation Introduction

## (a)Interpretation:The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is to be drawn.Concept Introduction:The molarity of a solution is the molar concentration of the solution; it measures the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as,M=nVWhere,n represents the number of moles of the solute. V represents the volume of the solution.The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The amount of a limiting reagent is lower than its required amount to complete the reaction.

Expert Solution

The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is represented as,

Figure 1.

### Explanation of Solution

The volume of 2.00M

copperII nitrate solution is 2.00L.

The volume of 3.00M potassium hydroxide solution is 2.00L.

The molarity of a solution is given as,

M=nV    (1)

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

Rearrange the above equation for the value of n.

n=MV    (2)

Substitute the value of molarity and volume of copperII nitrate solution in equation (2).

n=2.00M1molL11M2.00L=4.00mol

The number of moles of copperII nitrate present in the reaction mixture is 4.00mol.

Substitute the value of molarity and volume of potassium hydroxide solution in equation (2).

n=3.00M1molL11M2.00L=6.00mol

The number of moles of potassium hydroxide present in the reaction mixture is 6.00mol.

The balance chemical reaction between copperII nitrate and potassium hydroxide is represented as,

CuNO32aq+2KOHaqCuOH2s+2KNO3aq

The complete ionic reaction between copperII nitrate and potassium hydroxide is represented as,

Cu2+aq+2NO3aq+2K+aq+2OHaqCuOH2s+2K+aq+2NO3aq

One moles of CuNO32 reacts with two moles of KOH to produce one mole of CuOH2 and two moles of KNO3. Therefore, the relation between the number of moles of CuNO32 and KOH reacts is given as,

nCuNO32=nKOH2

Where,

• nKOH represents the number of moles of KOH.
• nCuNO32 represents the number of moles of CuNO32.

Substitute the value of number of moles of KOH in the above equation.

nCuNO32=6.00mol2=3.00mol

The number of moles of CuNO32 required to react with 6.00mol of KOH is 3.00mol but the number of moles of CuNO32 present in the reaction mixture is 4.00mol. Therefore, some amount of CuNO32 will remain unreacted in the reaction mixture. The ions that will remain in the solution even after the completion of the reaction are K+, Cu2+ and NO3. The picture of the solution made by mixing solution A and B together after the precipitation reaction takes place is represented as,

Figure 1.

Interpretation Introduction

### (b)Interpretation:The concentration of all ions left in solution after completion of the reaction and mass of the solid formed is to be calculated.Concept Introduction:The molarity of a solution is the molar concentration of the solution; it measures the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as,M=nVWhere,n represents the number of moles of the solute. V represents the volume of the solution.

Expert Solution

The concentration of Cu2+, NO3 and K+ ions left in solution after completion of the reaction between given solution of copperII nitrate and potassium hydroxide is 0.25M, 2.00M and 1.50M respectively. The mass of the precipitate formed after reaction between given solution of copperII nitrate and potassium hydroxide is 292.68g.

### Explanation of Solution

The complete ionic reaction between copperII nitrate and potassium hydroxide is represented as,

Cu2+aq+2NO3aq+2K+aq+2OHaqCuOH2s+2K+aq+2NO3aq

The initial number of moles of copperII nitrate is 4.00mol.

The number of mole of copperII nitrate reacted is 3.00mol.

The number of moles of copperII nitrate remains unreacted in the solution is given as,

n=n1n2

Where,

• n1 represents the initial number of moles of copperII nitrate.
• n2 represents the number of moles of copperII nitrate reacted.

Substitute the value of n1 and n2 in the above equation.

n=4.00mol3.00mol=1.00mol

One mole of CuNO32 dissociates in one mole of Cu2+ ion and two moles of NO3 ion. The number of moles of Cu2+ remains in the solution is 1.00mol.

The nitrate ions are not precipitated in the reaction therefore, the number of moles of nitrate ion present in the reaction before the reaction started and after the reaction completed will remain same. The relation between the number of moles of CuNO32 and NO3 ion is given as,

nNO3=2nCuNO32

Where,

• nNO3 represents the number of moles of NO3.
• nCuNO32 represents the number of moles of CuNO32.

Substitute the value of nCuNO32 in the above equation.

nNO3=24.00mol=8.00mol

The number of moles of NO3 ions present in the solution after completion of the reaction is 8.00mol.

The K+ ions are also not precipitated in the reaction. Therefore, the number of moles of K+ present in the reaction mixture is 6.00mol.

The volume of 2.00M

copperII nitrate solution is 2.00L.

The volume of 3.00M potassium hydroxide solution is 2.00L.

The total volume of the reaction mixture is given as,

V=V1+V2

Where,

• V1 represents the volume of copperII nitrate solution.
• V2 represents the volume of potassium hydroxide solution.

Substitute the value of V1 and V2 in the above equation.

V=2.00L+2.00L=4.00L

The molarity of a solution is given as,

M=nV    (1)

Where,

• n represents the number of moles of the solute.
• V represents the volume of the solution.

Substitute the value of number of moles of Cu2+ ion and total volume of the reaction mixture in the equation (1).

M=1.00mol4.00L1M1molL1=0.25M

Therefore, the concentration of Cu2+ ions left in solution after completion of the reaction between given solution of copperII nitrate and potassium hydroxide is 0.25M.

Substitute the value of number of moles of nitrate ion and total volume of the reaction mixture in the equation (1).

M=8.00mol4.00L1M1molL1=2.00M

Therefore, the concentration of nitrate ions left in solution after completion of the reaction between given solution of copperII nitrate and potassium hydroxide is 2.00M.

Substitute the value of number of moles of K+ ion and total volume of the reaction mixture in the equation (1).

M=6.00mol4.00L1M1molL1=1.50M

Therefore, the concentration of K+ ions left in solution after completion of the reaction between given solution of copperII nitrate and potassium hydroxide is 1.50M.

The limiting agent of the reaction is KOH. The amount of KOH available will decide the amount of CuOH2 produce.

The relation between the amount of number of moles of KOH reacted and number of moles of CuOH2 produced is given as,

nCuOH2=nKOH2

Where,

• nCuOH2 represents the number of moles of CuOH2.
• nKOH represents the number of moles of KOH.

Substitute the value of the available number of moles of KOH in the above equation.

nMgOH2=6.00mol2=3.00mol

The number of CuOH2 produced on mixing given solutions of copperII nitrate and potassium hydroxide is 3.00mol.

The molar mass of CuOH2 is 97.56gmol1.

The relation between the number of moles and mass of a substance is given as,

m=nMm

Where,

• n represents the number of the substance.
• Mm represents the molar mass of the substance.

Substitute the value of n and Mm in the above equation for the mass of CuOH2 produced.

m=3.00mol97.56gmol1=292.68g

Therefore, the mass of the precipitate formed after reaction between given solution of copperII nitrate and potassium hydroxide is 292.68g.

### Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!