Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.57P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 0 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(a)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 0.00 mL titrant and addition of 20.00 mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 11.86

Explanation of Solution

0mLHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of H+

Added amount of hydrogen ions H+ is as follows,

Initial number of moles of trimethylamine (CH3CH2)3N,

=mole(M)×Volume(V)=(0.1000mol(CH3CH2)3N/L)(103L/1mL)(20.00mL)=2.00×10-3mol(CH3CH2)3N 

Since no acid has been added, only the weak base (Kb) is important,

Determination the base ion concentration from the Kb and then determine the pOH from the pH.

Ka=5.2×104=[(CH3CH2)3NH+][OH][(CH3CH2)3N]=[x][x][0.1000x]=x2[0.1000]Solvefor(x),[OH]=x=7.2111×10-3M

Calculation for pOH

pOH=log[OH]=log(7.2111×10-3)pOH=2.141998

Therefore, the given solution pOH value is 2.14

Calculation for pH

pH=14.00pOH=14.002.141998=11.8580=11.86

(b)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 10mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(b)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 20.00 mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 10.72

Explanation of Solution

Determine 10mLofHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(10.00mL)=1.000×10-3molHCl 

The HCl will react with an equal amount of the base, and 1.000×103mol(CH3CH2)3N will remain, an equal number of moles of 1.000×103mol(CH3CH2)3NH+ will form.

The volume of the solution at this point is,

Volume

 =[(20.00+10.00)mL](103L/1mL)=0.03000L

Molarity of the excess (CH3CH2)3Nis,

=(1.000×103mol(CH3CH2)3N)/(0.03000L)=0.03333M

Molarity of the excess (CH3CH2)3NH+formed is,

=(1.000×103mol(CH3CH2)3NH+)/(0.03000L)=0.03333M

Since no acid has been added, only the weak base (Kb) is important,

Determination the base ion concentration from the Kb and then determine the pOH from pH.

Kb=5.2×104=[(CH3CH2)3NH+][OH][(CH3CH2)3N]=x(0.0333+x)(0.0333x)=x(0.03333)0.03333Solvefor(x),[OH]=x=5.2×10-4M

Calculation for pOH

pOH=log[OH]=log(5.2×10-4)pOH=3.283997

Therefore, the given solution pOH value is 3.28

Calculation for pH

pH=14.00pOH=14.003.283997=10.7160=10.72

(c)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 15 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(c)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 15.00 mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 10.24.

Explanation of Solution

Determine 15mLofHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(15.00mL)=1.500×10-3molHCl 

The HCl will react with an equal amount of the base, and 5.00×104mol(CH3CH2)3N will remain, an equal number of moles of 1.500×103mol(CH3CH2)3NH+ will form.

The volume of the solution at this point is,

Volume

 =[(20.00+10.00)mL](103L/1mL)=0.03500L

Molarity of the excess (CH3CH2)3Nis,

=(5.00×104mol(CH3CH2)3N)/(0.03500L)=0.0142857M

Molarity of the excess (CH3CH2)3NH+formed is,

=(1.500×103mol(CH3CH2)3NH+)/(0.03500L)=0.0428571M

Since no acid has been added, only the weak base (Kb) is important,

Determination the base ion concentration from the Kb and then determine the pOH from the pH.

Kb=5.2×104=[(CH3CH2)3NH+][OH][(CH3CH2)3N]=x(0.0428571+x)(0.0142857x)=x(0.0428571)0.0142857Solvefor(x),[OH]=x=1.7333×10-4M

Calculation for pOH

pOH=log[OH]=log(1.7333×10-4)pOH=3.761126

The given solution pOH value is 3.76

Calculation for pH

pH=14.00pOH=14.003.761126=10.23887=10.24

(d)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 19 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(d)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 19.00 mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 9.44

Explanation of Solution

Determine for 19.00mLHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(19.00mL)=1.900×10-3molHCl 

The HCl will react with an equal amount of the base, and 1.00×104mol(CH3CH2)3N will remain, an equal number of moles of 1.900×103mol(CH3CH2)3NH+ will form.

The volume of the solution at this point is,

Volume

 =[(20.00+19.00)mL](103L/1mL)=0.03900L

Molarity of the excess (CH3CH2)3Nis,

=(1.00×104mol(CH3CH2)3N)/(0.03900L)=0.002564103M

Molarity of the excess (CH3CH2)3NH+formed is,

=(1.900×103mol(CH3CH2)3NH+)/(0.03900L)=0.0487179M

Determination the base ion concentration from the Kb and then determine the pOH from the pH.

Kb=5.2×104=[(CH3CH2)3NH+][OH][(CH3CH2)3N]=x(0.0487179+x)(0.002564103x)=x(0.0487179)0.002564103Solvefor(x),[OH]=x=2.73684×10-5M

Calculation for pOH

pOH=log[OH]=log(2.73684×10-5)pOH=4.56275

The given solution pOH value is 4.57

Calculation for pH

pH=14.00pOH=14.004.56275=9.43725=9.44

(e)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 19.95 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(e)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 19.95mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 8.1

Explanation of Solution

Determine for 19.00mLHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(19.95mL)=1.995×10-3molHCl 

The HCl will react with an equal amount of the base, and 5.00×1046mol(CH3CH2)3N will remain, an equal number of moles of 1.995×103mol(CH3CH2)3NH+ will form.

The volume of the solution at this point is,

Volume

 =[(20.00+19.95)mL](103L/1mL)=0.03995L

Molarity of the excess (CH3CH2)3Nis,

=(5.00×106mol(CH3CH2)3N)/(0.03995L)=0.000125156M

Molarity of the excess (CH3CH2)3NH+formed is,

=(1.995×103mol(CH3CH2)3NH+)/(0.03995L)=0.0499374M

Determination the base ion concentration from the Kb and then determine the pOH from the pH.

Kb=5.2×104=[(CH3CH2)3NH+][OH][(CH3CH2)3N]=x(0.0499374+x)(0.000125156x)=x(0.0499374)0.000125156Solvefor(x),[OH]=x=1.303254×10-6M

Calculation for pOH

pOH=log[OH]=log(1.303254×10-6)pOH=5.88497

The given solution pOH value is 5.90

Calculation for pH

pH=14.00pOH=14.005.88497=8.11503=8.1

(f)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 20.00 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(f)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 20.00mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 6.01

Explanation of Solution

Determine for 20.00mLHClis added:

Trimethylamine is a weak base, when HCl is added the H+ concentration is calculated using the following formula.

pH=pKa+log[conjugate base][acid]

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(20.00mL)=2.000×10-3molHCl 

The HCl will react with an equal amount of the base, and 0mol(CH3CH2)3N will remain, and 2.000×103molof(CH3CH2)3NH+ will form. This is the equivalent point. 

The volume of the solution at this point is,

Volume

 =[(20.00+20.00)mL](103L/1mL)=0.04000L

Molarity of the excess (CH3CH2)3NH+formed is,

=(2.000×103mol(CH3CH2)3NH+)/(0.04000L)=0.05000M

CalculateKafor (CH3CH2)3NH+

Kb=KW/Ka=(1.0×1014)(5.2×104)=1.9231×1011

Determination the base ion concentration from the Ka and then determine the pH from the pOH.

Kb=1.9231×1011=[H3O+][(CH3CH2)3N][(CH3CH2)3NH+]=[x][x](0.05000x)=[x][x][0.05000]Solvefor(x),[H3O+]=x=9.80587×10-7M

Calculation for pH

pH=log[H3O+]=log(9.80587×10-7)pH=6.0085

The given solution pH value is 6.01

(g)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 20.05 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(g)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 20.05mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 3.90.

Explanation of Solution

Determine for 20.05mLHClis added:

After the equivalence point, the excess strong acid is the primary factor influencing the pH.

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(20.05mL)=2.005×10-3molHCl 

The HCl will react with an equal amount of the base, and 0mol(CH3CH2)3N will remain, and 5.0×106molof(CH3CH2)3NH+ will be in excess.

The volume of the solution at this point is,

Volume

 =[(20.00+20.05)mL](103L/1mL)=0.04005L

Molarity of the excess H3O+ is,

=(5.0×106molH3O+)/(0.04005L)=1.248×104M

Determination the base ion concentration from the Ka and then determine the pH from the pOH.

Calculation for pH

pH=log[H3O+]=log(1.248×104)pH=3.9036

The given solution pH value is 3.90

(h)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl solution after the addition of 25.00 mL titrant should be calculated.

Concept introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid]

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

(h)

Expert Solution
Check Mark

Answer to Problem 19.57P

The pH of a given buffer solution after the addition 10.00 mL titrant and addition of 25.00mL and 0.1000 M trimethylamine, (Kb=5.2×104 ) with 0.1000 M HCl is 1.95

Explanation of Solution

Determine for 25.00mLHClis added:

Determine the moles of HCl is added, 

=mole(M)×Volume(V)=(0.1000molHCl/L)(103L/1mL)(25.00mL)=2.500×10-3molHCl 

The HCl will react with an equal amount of the base, and 0mol(CH3CH2)3N will remain, and 5.00×104molof(CH3CH2)3NH+ will be in excess.

The volume of the solution at this point is,

Volume

 =[(20.00+25.00)mL](103L/1mL)=0.04500L

Molarity of the excess H3O+ is,

=(5.00×104molH3O+)/(0.04500L)=1.1111×102M

Calculation for pH

pH=log[H3O+]=log(1.1111×102)pH=1.9542

The given solution pH value is 1.95.

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Chapter 19 Solutions

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - The scenes below depict solutions of the same...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.44PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.84PCh. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.88PCh. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.96PCh. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.100PCh. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.105PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. Two...Ch. 19 - It is possible to detect NH3 gas over 10−2 M NH3....Ch. 19 - Manganese(II) sulfide is one of the compounds...Ch. 19 - The normal pH of blood is 7.40 ± 0.05 and is...Ch. 19 - A bioengineer preparing cells for cloning bathes a...Ch. 19 - Sketch a qualitative curve for the titration of...Ch. 19 - A solution contains 0.10 M ZnCl2 and 0.020 M...Ch. 19 - Prob. 19.120PCh. 19 - The scene at right depicts a saturated solution of...Ch. 19 - Prob. 19.122PCh. 19 - The acid-base indicator ethyl orange turns from...Ch. 19 - Prob. 19.124PCh. 19 - Prob. 19.125PCh. 19 - Prob. 19.126PCh. 19 - Prob. 19.127PCh. 19 - Prob. 19.128PCh. 19 - Prob. 19.129PCh. 19 - Calcium ion present in water supplies is easily...Ch. 19 - Calculate the molar solubility of Hg2C2O4 (Ksp =...Ch. 19 - Environmental engineers use alkalinity as a...Ch. 19 - Human blood contains one buffer system based on...Ch. 19 - A 0.050 M H2S solution contains 0.15 M NiCl2 and...Ch. 19 - Quantitative analysis of Cl− ion is often...Ch. 19 - An ecobotanist separates the components of a...Ch. 19 - Some kidney stones form by the precipitation of...Ch. 19 - Prob. 19.138PCh. 19 - Prob. 19.139PCh. 19 - Because of the toxicity of mercury compounds,...Ch. 19 - A 35.0-mL solution of 0.075 M CaCl2 is mixed with...Ch. 19 - Rainwater is slightly acidic due to dissolved CO2....Ch. 19 - Prob. 19.143PCh. 19 - Ethylenediaminetetraacetic acid (abbreviated...Ch. 19 - Buffers that are based on...Ch. 19 - NaCl is purified by adding HCl to a saturated...Ch. 19 - Scenes A to D represent tiny portions of 0.10 M...Ch. 19 - Prob. 19.148PCh. 19 - Prob. 19.149PCh. 19 - Prob. 19.150P
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