EBK BIOCHEMISTRY
EBK BIOCHEMISTRY
6th Edition
ISBN: 9781337431200
Author: GRISHAM
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 2, Problem 1P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Calculating pH from [H+] Calculate the pH of the following.

  1. 5 ×10-4 M HCl
  2. 7 ×10-5 M NaOH
  3. M HCl
  4. 7 ×10-2 M KOH
  5. 04 m HCl
  6. 6 ×10-9 M HCl

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given acid concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Answer to Problem 1P

pH =3.3

Explanation of Solution

Given information:

HCl concentration =5×104M

Formula used:

  pH=log[H+]

  [H+]= Hydrogen ion concentration in moles per liter

Calculation:

Since, HCl is a strong acid, the concentration of hydrogen ion is equal to the concentration of the acid.

  pH=log[H+]=log(5×104)=3.3

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given base concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Similarly, for a base, pOH can be calculated by taking negative log of the hydroxide ion concentration.

  pOH=log[OH]

The relation between pH and pOH is as follows:

  pH=14pOH

Answer to Problem 1P

pH =9.85

Explanation of Solution

Given information:

NaOH concentration =7×105M

Formula used:

  pOH=-log[OH-]

  [H+]= Hydrogen ion concentration in moles per liter

  pH+pOH=14

Calculation:

Since, NaOH is a strong base, the concentration of hydroxide ion will be equal to the concentration of the NaOH.

  pOH=-log[OH-]=-log(7×10-5)=4.15pH=14-4.15=9.85

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given acid concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Answer to Problem 1P

pH =5.7

Explanation of Solution

Given information:

HCl concentration =2×106M

Formula used:

  pH=log[H+]

  [H+]= Hydrogen ion concentration in moles per liter

Calculation:

Since, HCl is a strong acid, the concentration of hydrogen ion is equal to the concentration of the acid.

  pH=log[H+]=log(2×106)=5.7

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given base concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Similarly, for a base, pOH can be calculated by taking negative log of the hydroxide ion concentration.

  pOH=log[OH]

The relation between pH and pOH is as follows:

  pH=14pOH

Answer to Problem 1P

pH =12.5

Explanation of Solution

Given information:

KOH concentration =3×102M

Formula used:

  pOH=log[OH]

  [H+]= Hydrogen ion concentration in moles per liter

  pH+pOH=14

Calculation:

Since, KOH is a strong base, the concentration of hydroxide will be the concentration of KOH.

  pOH=-log[OH-]=-log(3×10-2)=1.52pH=14-1.52=12.5

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given acid concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Answer to Problem 1P

pH =4.4

Explanation of Solution

Given information:

HCl concentration =0.04×103M

Formula used:

  pH=log[H+]

  [H+]= Hydrogen ion concentration in moles per liter

Calculation:

Since, HCl is a strong acid, the concentration of hydrogen ion is equal to the concentration of the acid.

  pH=log[H+]=log(0.04×103)=4.4

Expert Solution
Check Mark
Interpretation Introduction

To determine:

The pH value of given acid concentration.

Introduction:

The pH is the measure of hydrogen ion concentration in a solution. pH value is calculated by taking the negative base 10 logarithm of the hydrogen ion molar concentration. When the hydrogen ion concentration is high, the pH value is low and when the hydrogen ion concentration is low, pH value is high.

It is mathematically represented as follows:

  pH=log[H+]

Here, H+is hydrogen ion concentration.

Answer to Problem 1P

pH =6.99

Explanation of Solution

Given information:

HCl concentration =6×109M

Formula used:

  pH=log[H+]

  [H+]= Hydrogen ion concentration in moles per liter

Calculation:

Here, ion production of water is considered,

Equation 1: [H+]+[OH]=kw=1014

In the solution sum of positive charges is equal to the sum of negative charges,

Equation 2: [H+]=[Cl]+[OH]

HCl is fully dissociated,

Equation 3: [Cl]=[HCltotal]=6×109

Substitute equation 2 to equation 3, solve equation 2 to get [OH-] and substitute to equation 1 to get [H+],

  [H+]26×109[H+]1014=0

  [H+]=b±b24ac2a=6×109±(6×109)2+4×10142[H+]=6×109+(6×109)2+4×10142=1.03×107

  pH=log[H+]=log(1.03×107)=6.99

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