General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 4, Problem 4.117SP
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=MassofthecompoundMolarmassofthecompound

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
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Answer to Problem 4.117SP

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

Concentrations of Na+ = 0.5295M

Concentrations of OH- = 0.09968M

Concentrations of NO3+ = 0.4298M

Concentrations of Mg2+ = 0

Record the given data.

Volume of Mg(NO3)2 solution = 22.02mL

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)2mole=1.615g×1moleMg(NO3)2148.33gMg(NO3)2=0.010888moleMg(NO3)2

Molar mass of NaOH=39.998 g

NaOHmole=1.073gNaOH×1moleNaOH39.998gNaOH=0.026826moleNaOH

  • 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 0.010888moles
  • The mole of NaOH is 0.026826moles

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

Concentration of Na+ is,

[Na+]=0.026826moleNaOH×1moleNa+1molNaOH×10.05066L=0.5295M

Concentration of NO3- is,

[NO3-]=0.010888moleMg(NO3)2×2moleNO3-1moleMg(NO3)2×10.05066L=0.4298M

Concentration of OH- reacted with Mg2+ is,

[OH-]=0.010888mloeMg2+×2moleOH-1moleMg2+=0.021776moleOH-

Concentration of excess OH- is,

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

Concentration of excess Mg2+ is,

[Mg2+]=0M

  • 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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Chapter 4 Solutions

General Chemistry

Ch. 4.5 - Prob. 1PECh. 4.5 - Prob. 2PECh. 4.5 - Prob. 3PECh. 4.5 - Prob. 1RCCh. 4.6 - Prob. 1PECh. 4.6 - Prob. 1RCCh. 4.6 - Prob. 2PECh. 4.6 - Prob. 3PECh. 4 - Prob. 4.1QPCh. 4 - Prob. 4.2QPCh. 4 - Prob. 4.3QPCh. 4 - 4.4 What is the difference between the following...Ch. 4 - 4.5 Water is an extremely weak electrolyte and...Ch. 4 - Prob. 4.6QPCh. 4 - Prob. 4.7QPCh. 4 - 4.8 Which of the following diagrams best...Ch. 4 - Prob. 4.9QPCh. 4 - Prob. 4.10QPCh. 4 - Prob. 4.11QPCh. 4 - Prob. 4.12QPCh. 4 - Prob. 4.13QPCh. 4 - Prob. 4.14QPCh. 4 - Prob. 4.15QPCh. 4 - Prob. 4.16QPCh. 4 - Prob. 4.17QPCh. 4 - Prob. 4.18QPCh. 4 - Prob. 4.19QPCh. 4 - Prob. 4.20QPCh. 4 - 4.21 Write ionic and net ionic equations for the...Ch. 4 - Prob. 4.22QPCh. 4 - Prob. 4.23QPCh. 4 - Prob. 4.24QPCh. 4 - Prob. 4.25QPCh. 4 - Prob. 4.26QPCh. 4 - Prob. 4.27QPCh. 4 - Prob. 4.28QPCh. 4 - Prob. 4.29QPCh. 4 - Prob. 4.30QPCh. 4 - Prob. 4.31QPCh. 4 - Prob. 4.32QPCh. 4 - Prob. 4.33QPCh. 4 - Prob. 4.34QPCh. 4 - Prob. 4.35QPCh. 4 - Prob. 4.36QPCh. 4 - Prob. 4.37QPCh. 4 - Prob. 4.38QPCh. 4 - 4.39 For the complete redox reactions given here,...Ch. 4 - Prob. 4.40QPCh. 4 - Prob. 4.41QPCh. 4 - Prob. 4.42QPCh. 4 - Prob. 4.43QPCh. 4 - Prob. 4.44QPCh. 4 - Prob. 4.45QPCh. 4 - Prob. 4.46QPCh. 4 - Prob. 4.47QPCh. 4 - Prob. 4.48QPCh. 4 - Prob. 4.49QPCh. 4 - Prob. 4.50QPCh. 4 - Prob. 4.51QPCh. 4 - Prob. 4.52QPCh. 4 - Prob. 4.53QPCh. 4 - Prob. 4.54QPCh. 4 - Prob. 4.55QPCh. 4 - Prob. 4.56QPCh. 4 - Prob. 4.57QPCh. 4 - Prob. 4.58QPCh. 4 - Prob. 4.59QPCh. 4 - Prob. 4.60QPCh. 4 - Prob. 4.61QPCh. 4 - Prob. 4.62QPCh. 4 - Prob. 4.63QPCh. 4 - Prob. 4.64QPCh. 4 - Prob. 4.65QPCh. 4 - Prob. 4.66QPCh. 4 - Prob. 4.67QPCh. 4 - Prob. 4.68QPCh. 4 - Prob. 4.69QPCh. 4 - 4.70 Distilled water must be used in the...Ch. 4 - 4.71 If 30.0 mL of 0.150 M CaCl2 is added to 15.0...Ch. 4 - Prob. 4.72QPCh. 4 - Prob. 4.73QPCh. 4 - Prob. 4.74QPCh. 4 - Prob. 4.75QPCh. 4 - Prob. 4.76QPCh. 4 - Prob. 4.77QPCh. 4 - Prob. 4.78QPCh. 4 - Prob. 4.79QPCh. 4 - Prob. 4.80QPCh. 4 - Prob. 4.81QPCh. 4 - Prob. 4.82QPCh. 4 - Prob. 4.83QPCh. 4 - Prob. 4.84QPCh. 4 - Prob. 4.85QPCh. 4 - Prob. 4.86QPCh. 4 - Prob. 4.87QPCh. 4 - Prob. 4.88QPCh. 4 - Prob. 4.89QPCh. 4 - Prob. 4.90QPCh. 4 - Prob. 4.91QPCh. 4 - Prob. 4.92QPCh. 4 - Prob. 4.93QPCh. 4 - 4.74 The molecular formula of malonic acid is...Ch. 4 - Prob. 4.95QPCh. 4 - Prob. 4.96QPCh. 4 - Prob. 4.97QPCh. 4 - Prob. 4.98QPCh. 4 - Prob. 4.99QPCh. 4 - Prob. 4.100QPCh. 4 - Prob. 4.101QPCh. 4 - Prob. 4.102QPCh. 4 - 4.103 These are common household compounds: table...Ch. 4 - Prob. 4.104QPCh. 4 - Prob. 4.105QPCh. 4 - Prob. 4.106QPCh. 4 - 4.107 A number of metals are involved in redox...Ch. 4 - Prob. 4.108QPCh. 4 - Prob. 4.109QPCh. 4 - Prob. 4.110QPCh. 4 - Prob. 4.111QPCh. 4 - Prob. 4.112QPCh. 4 - Prob. 4.114SPCh. 4 - Prob. 4.115SPCh. 4 - Prob. 4.116SPCh. 4 - Prob. 4.117SPCh. 4 - Prob. 4.118SP
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