Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 4, Problem 154CP

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution:

IO3(aq) + I(aq) → I3(aq)

Triiodide ion concentration is determined by titration with a sodium thiosulfate (Na2S2O3) solution. The products are iodide ion and tetrathionate ion (S4O6).

a. Balance the equation for the reaction of IO3 with I ions.

b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HQ required to convert all of the IO3 ions to I ions?

c. Write and balance the equation for the reaction of S2O32− with I3 in acidic solution.

d. A 25.00-mL sample of a 0.0100 M solution of KIO. is reacted with an excess of KI. It requires 32.04 mL of Na2S2O3 solution to titrate the I3 ions present. What is the molarity of the Na2S2O3 solution?

e. How would you prepare 500.0 mL of the KIO3 solution in part d using solid KIO3?

(a)

Expert Solution
Check Mark
Interpretation Introduction

The balanced equation for the given reaction, minimum mass of KI and minimum volume of 3M HCl required for the reaction, balanced equation for the reaction between S2O32- and I3, molarity of Na2S2O3 and preparation of KIO3 in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

  • Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
  • Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
  • Club the both half-reactions.
  • Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
  • Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

  • Equation for finding molarity of a solution is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Answer to Problem 154CP

The balanced equations for the given reaction is,

IO3(aq)+8I(aq)+6H+(aq)3I3(aq)+3H2O(aq)

Explanation of Solution

To determine: The balanced equations for the given reactions.

The chemical equation for the given redox reaction is,

IO3(aq)+I(aq)I3(aq)

Steps (I, II and III) followed for balance the redox equation,

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

I(aq)I3(aq)

Reduction reaction is,

IO3(aq)I3(aq)

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

3I(aq)I3(aq)+2e

18H+(aq)+3IO3(aq)+16eI3(aq)+9H2O(aq)

Equalizing the numbers of electrons in oxidation and reduction reactions,

24I(aq)8I3(aq)+16e

III- club the both half-reactions into a single equation,

24I(aq)+18H+(aq)+3IO3(aq)+16eI3(aq)+9H2O(aq)+8I3(aq)+16e

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

24I(aq)+18H+(aq)+3IO3(aq)9I3(aq)+9H2O(aq)IO3(aq)+8I(aq)+6H+(aq)3I3(aq)+3H2O(aq)

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of KI and minimum volume of 3M HCl required for the reaction, balanced equation for the reaction between S2O32- and I3, molarity of Na2S2O3 and preparation of KIO3 in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

  • Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
  • Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
  • Club the both half-reactions.
  • Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
  • Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

  • Equation for finding molarity of a solution is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Answer to Problem 154CP

The minimum mass of KI and minimum volume of 3M HCl required for the reaction is 3.7315g and 0.7492×10-2L.

Explanation of Solution

To determine: the minimum mass of KI and minimum volume of 3M HCl required for the reaction if 0.6013g of KIO3 is used.

The mass of KIO3 in the reaction is given as 0.6013g.

Equation for number of moles of a substance is,

numberofmoles=givenmassmolarmass

Therefore,

The number of moles of KIO3 is,

numberofmoles=0.6013g214g=0.28098×10-2mol

The balanced equations for the given reaction is,

IO3(aq)+8I(aq)+6H+(aq)3I3(aq)+3H2O(aq)

The mole ratio between IO3 and I is 1:8.

Therefore,

The number of moles of I reacted with 0.28098×10-2mol of IO3 is,

8×0.28098×10-2mol=0.022478mol

Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

Therefore,

The minimum mass of KI required in the reaction is,

0.022478mol×166.0028g=3.7315g

The number of moles of HCl is equals to the number of mole of KI, that is 0.022478mol.

The volume of 3M HCl reacted can be find out from its molarity and number of moles,

That is,

volume of solution(L)=0.022478mol3mol/L=0.7492×10-2L

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of KI and minimum volume of 3M HCl required for the reaction, balanced equation for the reaction between S2O32- and I3, molarity of Na2S2O3 and preparation of KIO3 in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

  • Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
  • Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
  • Club the both half-reactions.
  • Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
  • Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

  • Equation for finding molarity of a solution is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Answer to Problem 154CP

The balanced equation for the reaction between S2O32- and I3 is,

Explanation of Solution

To determine: the balanced equations for the reaction between S2O32- and I3.

The chemical equation for the given redox reaction is,

S2O32-(aq)+I3(aq)S4O62-(aq)+I2(aq)

Steps (I, II and III) followed for balance the redox equation,

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

S2O32-(aq)S4O62-(aq)

Reduction reaction is,

I3(aq)I2(aq)

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

2S2O32-(aq)S4O62-(aq)+2e

2I3(aq)+1e3I2(aq)

Equalizing the numbers of electrons in oxidation and reduction reactions,

4I3(aq)+2e6I2(aq)

III- club the both half-reactions into a single equation,

2S2O32-(aq)+4I3(aq)+2eS4O62-(aq)+2e+6I2(aq)

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

2S2O32-(aq)+4I3(aq)S4O62-(aq)+6I2(aq)

To determine: the molarity of Na2S2O3 in the above reaction if 25mL of 0.01M KIO3 is reacted with 32.04mL of Na2S2O3.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of KI and minimum volume of 3M HCl required for the reaction, balanced equation for the reaction between S2O32- and I3, molarity of Na2S2O3 and preparation of KIO3 in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

  • Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
  • Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
  • Club the both half-reactions.
  • Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
  • Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

  • Equation for finding molarity of a solution is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Answer to Problem 154CP

The molarity of Na2S2O3 is 0.375×10-3mol.

Explanation of Solution

The molarity of KIO3 in the reaction is given as 0.01M.

The volume of KIO3 in the reaction is given as 25mL.

Equation for number of moles of a substance from its molarity and volume is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Therefore,

The number of moles of KIO3 is,

numberofmoles=0.01mol/L×25×10-3L=0.25×10-3mol

The balanced equations for the reaction of IO3 and product I3 is,

IO3(aq)+8I(aq)+6H+(aq)3I3(aq)+3H2O(aq)

The mole ratio between IO3 and I3 is 1:3.

Therefore,

The number of moles of I3 produced with 0.25×10-3mol of IO3 is,

3×0.25×10-3mol=0.75×10-3mol

The balanced equations for the reaction between S2O32- and I3 is,

2S2O32-(aq)+4I3(aq)S4O62-(aq)+6I2(aq)

The mole ratio between S2O32- and I3 is 1:2.

Therefore,

The number of moles of S2O32- reacted with 0.75×10-3mol of I3 is,

12×0.75×10-3mol=0.375×10-3mol

To explain: the preparation of KIO3 solution from solid in part (d).

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of KI and minimum volume of 3M HCl required for the reaction, balanced equation for the reaction between S2O32- and I3, molarity of Na2S2O3 and preparation of KIO3 in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

  • Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
  • Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
  • Club the both half-reactions.
  • Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
  • Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

  • Equation for finding molarity of a solution is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Answer to Problem 154CP

The preparation of KIO3 in part d is by dissolving 1.07g of solid KIO3 in water.

Explanation of Solution

The molarity of KIO3 in the reaction is given as 0.01M.

The volume of KIO3 is to prepare is 500mL=0.5L.

Equation for number of moles of a substance from its molarity and volume is,

Molarity(M)=number of moles of solute(mol)volume of solution(L)

Therefore,

The number of moles of KIO3 to be prepared is,

numberofmoles=0.01mol/L×0.5L=0.005mol

Equation for number of grams of a substance from its number of moles is,

Number of moles×Molecularmass in grams=Numberofgrams

Therefore,

The mass of solid KIO3 to be prepared is,

0.005×214g=1.07g

Dissolving 1.07g of KIO3 in water of 500mL will give the solution of part d.

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Chapter 4 Solutions

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