   # Consider the titration of 100.0 mL of 0.200 M HONH 2 by 0.100 M HCI. (K b for HONH 2 = 1.1 × 10 −8 .) a. Calculate the pH after 0.0 mL of HCl has been added. b. Calculate the pH after 25.0 mL of HCl has been added. c. Calculate the pH after 70.0 mL of HCl has been added. d. Calculate the pH at the equivalence point. e. Calculate the pH after 300.0 mL of HCl has been added. f. At what volume of HCl added does the pH = 6.04? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 106CWP
Textbook Problem
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## Consider the titration of 100.0 mL of 0.200 M HONH2 by 0.100 M HCI. (Kb for HONH2 = 1.1 × 10−8.)a. Calculate the pH after 0.0 mL of HCl has been added.b. Calculate the pH after 25.0 mL of HCl has been added.c. Calculate the pH after 70.0 mL of HCl has been added.d. Calculate the pH at the equivalence point.e. Calculate the pH after 300.0 mL of HCl has been added.f. At what volume of HCl added does the pH = 6.04?

(a)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The pH of the solution after the addition of 0.0mL of HCl.

### Explanation of Solution

Explanation

Given

Titration of 100.0ml of 0.200MHONH2 with 0.100MHCl.

The base dissociation constant Kb is 1.1×108.

The given solution is the combination of acid and a base. The acid-base reaction for this combination is given as,

HONH2+HClNH2Cl+H2O

When 0.0mlKOH is added to the solution it means only HONH2 is present in the solution.

Therefore the equilibrium of HONH2 is represented by the equation,

HONH2NH2++OH-

Molarity of HONH2 is 0.200M. The number of moles is calculated by the formula,

n=C×V

Where,

• n is the number of moles.
• C is the concentration of the solution.
• V is the volume of the solution.

Substitute the values of concentrations and volume in the above equation.

n=C×V=0.200M×1001000L=0.02mol

Now x is supposed to be the change in moles. The equilibrium reaction with the calculated moles is expressed in ICE (initial, change, equilibrium) table as,

HONH2NH2++OH-Initialmol0.0200Changeinmolx+x+xEquilibriummol0.02x+x+x

The base dissociation constant (Kb) for this reaction is written as,

Kb=(x)(x)(0

(b)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The pH of the solution after the addition of 25.0mL of HCl.

(c)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The pH of the solution after the addition of 70.0mL of KOH.

(d)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The pH of the solution at the equivalence point.

(e)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The pH of the solution after the addition of 300.0mL of HCl.

(f)

Interpretation Introduction

Interpretation: The pH of the solutions after the addition of different volumes of 0.100M HCl are to be calculated.

Concept introduction: HONH2 is a weak base and HCl is a strong acid. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as pH of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

To determine: The volume of HCl added to make the pH-6.04.

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