The concentrations of ions present in mixture of NaOH and Mg(NO 3 ) 2 solutions should be calculated. Concept introduction: Precipitation reaction: If precipitate is formed, when two solutions are mixed together is known as precipitation reaction. The mass of the precipitate is depends on the reactant masses. Molarity: The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution. Molarity = No .mole volume (L) Mole: The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound. Mole = Massof the compound Molar mass of the compound Mole: The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is, Mole = Concentration (M)× volume (L)

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Answer 1

Question

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

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Answer 2

Question

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

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Answer 3

Question

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

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Answer 4

Question

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

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Answer 5

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

Answer 6

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.

Conclusion

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions were calculated.

Answer 7

Chapter 4, Problem 4.165QP

Interpretation Introduction

Interpretation:

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two solutions are mixed together is known as precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$Molarity = No.molevolume (L)$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$Mole = Massof the compoundMolar mass of the compound$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$Mole = Concentration (M)× volume (L)$

Answer 8

Expert Solution & Answer

Answer to Problem 4.165QP

The concentrations of ions present in mixture of $NaOH$ and $Mg(NO3)2$ solutions are,

Concentrations of $Na+$ = $0.5295 M$

Concentrations of $OH-$ = $0.09968 M$

Concentrations of $NO3+$ = $0.4298 M$

Concentrations of $Mg2+$ = $0$

Record the given data.

Volume of $Mg(NO3)2$ solution = $22.02 mL$

Volume of $NaOH$ solution = $28.64 mL$

Mass of $Mg(NO3)2$ in solution = $1.615 g$

Mass of $NaOH$ in solution = 1.073 g

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given volume and masses of $NaOH$ and $Mg(NO3)2$ their solutions are recorded as shown above.

To calculate the mole of $NaOH$ and $Mg(NO3)2$ in their solutions

The reaction of $Mg(NO3)2$ with $NaOH$ is,

$Mg(NO3)2(aq)+2NaOH(aq)→Mg(OH)2(s)↓+Na2(NO3)2(aq)$

Molar mass of $Mg(NO3)2$ = $148.33 g$

$Mg(NO3)2 mole = 1.615g ×1mole Mg(NO3)2148.33 g Mg(NO3)2 = 0.010888 mole Mg(NO3)2$

Molar mass of $NaOH$ = $39.998 g$

$NaOH mole = 1.073 gNaOH×1moleNaOH39.998 g NaOH = 0.026826 mole NaOH$

The mass of substance taken is divided by molar mass of substance to give the mole of the substance present in taken mass of substance.
The masses and molar masses of $NaOH$ and $Mg(NO3)2$ are plugged in above equations to give mole of $NaOH$ and $Mg(NO3)2$
The mole of $Mg(NO3)2$ is $0.010888 moles$
The mole of $NaOH$ is $0.026826 moles$

To calculate the moles of ions in present the solution mixture of $NaOH$ and $Mg(NO3)2$

Concentration of $Na+$ is,

$[Na+] = 0.026826 mole NaOH×1 mole Na+1 mol NaOH×1 0.05066 L = 0.5295 M$

Concentration of $NO3-$ is,

$[NO3-] = 0.010888 mole Mg(NO3)2×2 mole NO3-1 moleMg(NO3)2×10.05066 L = 0.4298 M$

Concentration of $OH-$ reacted with $Mg2+$ is,

$[OH-] = 0.010888 mloe Mg2+×2 mole OH- 1 mole Mg2+ = 0.021776 mole OH-$

Concentration of excess $OH-$ is,

$[OH-] excess= 0.26826 mole - 0.021776 mole = 0.005050 mole = 0.005050 mole0.05066 L = 0.09968 M$

Concentration of excess $Mg2+$ is,

$[Mg2+] = 0 M$

From the balanced equation the one mole of $Mg(NO3)2$ is required two mole of $NaOH$ to form a $Mg(OH)2$ precipitate. Therefore $Mg(NO3)2$ is a limiting reagent.
The Concentration of excess $OH-$ is calculated by subtract the calculated concentration of $OH-$ reacted with $Mg2+$ from total concentration of $OH-$ .
Moles of compounds and required number of moles of ions are plugged in the equation to give concentrations of ions present in the given solution mixture.
$Mg2+$ ions are completely react give a precipitate. So, the concentration of $Mg2+$ present in the mixture of solution is zero.