General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
Question
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Chapter 4, Problem 4.79QP

(a)

Interpretation Introduction

Interpretation:

The volume of NaOH solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

  • In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
  • In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

Mc×Vc=Md×VdMc=InitialconcentrationVc=InitialvolumeMd=FinalconcentrationVd=Finalvolume

(a)

Expert Solution
Check Mark

Answer to Problem 4.79QP

The volume of NaOH solution is needed to titrate is 42.78mL

Explanation of Solution

Equation for the reaction is

NaOH+HClNaCl+H2O

Initial concentration of the HCl solution Mc = 2.430 M

Initial volume of the HCl  solution Vc= 25 mL

Final concentration of the NaOH solution Md = 1.420M

The final volume of the NaOH solution can be calculated as,

Mc×Vc=Md×Vd

(2.430M)×(25mL)=(1.420M)×VdVd=(2.430M)×(25mL)(1.420M)=42.78mL

(b)

Interpretation Introduction

Interpretation:

The volume of NaOH solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

  • In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
  • In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

Mc×Vc=Md×VdMc=InitialconcentrationVc=InitialvolumeMd=FinalconcentrationVd=Finalvolume

(b)

Expert Solution
Check Mark

Answer to Problem 4.79QP

The volume of NaOH solution is needed to titrate is 158.4mL

Explanation of Solution

Equation for the reaction is

2NaOH+H2SO4Na2SO4+2H2O

Initial concentration of the H2SO4 solution Mc = 4.500 M

Initial volume of the H2SO4  solution Vc= 25 mL

Final concentration of the NaOH solution Md = 1.420M

The final volume of the NaOH solution can be calculated as,

Mc×Vc=Md×Vd

(4.500 M)×(25mL)=(1.420M)×VdVd=(4.500 M)×(25mL)(1.420M)=79.22mL

Since two moles of NaOH are required to titrate with one mole of H2SO4,

79.22mL×2=158.4mL

(c)

Interpretation Introduction

Interpretation:

The volume of NaOH solution is needed to titrate the given solution has to be determined.

Concept introduction:

Volumetric principle:

  • In the neutralization process, the volume and concentration of initial components are equal to the volume and concentration of the final components.
  • In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation and it is,

Mc×Vc=Md×VdMc=InitialconcentrationVc=InitialvolumeMd=FinalconcentrationVd=Finalvolume

(c)

Expert Solution
Check Mark

Answer to Problem 4.79QP

The volume of NaOH solution is needed to titrate is 79.22mL

Explanation of Solution

Equation for the reaction is

3NaOH+H3PO4Na3(PO4)+3H2O

Initial concentration of the H3PO4 solution Mc = 1.500 M

Initial volume of the H3PO4  solution Vc= 25 mL

Final concentration of the NaOH solution Md = 1.420M

The final volume of the NaOH solution can be calculated as,

Mc×Vc=Md×Vd

(1.500 M)×(25mL)=(1.420M)×VdVd=(1.500 M)×(25mL)(1.420M)=26.40mL

Since 3 moles of NaOH are required to titrate with one mole of H3PO4,

26.40mL×3=79.22mL

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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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