Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.56P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 0 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa:

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base.

(a)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 0.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 2.91.

Explanation of Solution

Given,

At 0mL of base added, the concentration of [H3O]+ is dependent on the dissociation of Butanoic acid.

ICE Table of equilibrium,

 H-But(aq) +H2O(aq)H3O+(l) +But(aq)Initial (M)0.100M   00Change(M)x+x+xEquilibrium  0.100xxx

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

  Ka=1.54×105=[H3O+][But][HBut]=[x][x][0.100x]=x2[0.1000][H3O+]=x=1.2409674×103M

Calculation for pH and pOH

  pH=log[H3O+]=log(1.2409674×103)pH=2.9062=2.91

(b)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 10.00 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid.

Base ionization constant Kb:

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base.

(b)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the 10.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 4.81.

Explanation of Solution

Given,

The initial number of moles of HBut,

=(M)(V)=(0.1000molHBut/L)(103L/1mL)(20.00mL)=2.00×10-3molHBut

Identification the moles of NaOH added

Moles added NaOH

=(0.1000molNaOH/L)(103L/1mL)(10.00mL)=1.000×103molNaOH

The NaOH will react with an equal amount of the aid, and 1.00×103molH-But will remain. An equal number of moles of But will form.

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol1.000×103mol   0Change(M)1.000×103mol1.000×103mol+1.000×103molEquilibrium  1.000×103mol01.000×103mol

Determination the liters of solution present at the equivalent point,

Volume

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

Concentration of NH4+ at equivalent point,

Molarity of the excess H-But

  =(1.000×103molH-But)/(0.03000L)=0.03333M

Molarity of the excess Butformed is,

  =(1.000×103molBut)/(0.03000L)=0.03333M

Using a reaction table for the equilibrium reaction of H-But

   H-But+H2OH3O+ +ButInitial (M):0.03333M  00.03333MChange(M):x+x+xEquilibrium: 0.03333Mx+x0.03333M+x

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

  Ka=1.54×105=[H3O+][But][HBut]=x(0.0333+x)(0.0333x)=x(0.03333)0.03333[H3O+]=x=1.54×105M

Calculation for pH and pOH

  pH=log[H3O+]=log(1.54×105)pH=4.812479=4.81

(c)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 15.00 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa:

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(c)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 15.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 5.29.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(15.00mL)=1.500×103molNaOH

The NaOH will react with an equal amount of the aid, and 5.00×104molH-But will remain. An equal number of moles of But will form.

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol1.500×103mol  0Change(M)1.500×103mol1.500×103mol+1.500×103molEquilibrium  5.000×104mol01.500×103mol

Determination the liters of solution present at the equivalent point,

Volume

   =[(20.00+15.00)mL](103L/1mL)=0.03500L

Concentration of NH4+ at equivalent point,

Molarity of the excess H-But

  =(5.000×104molH-But)/(0.03500L)=0.0142857M

Molarity of the excess Butformed is,

  =(1.500×103molBut)/(0.03500L)=0.042857M

Using a reaction table for the equilibrium reaction of H-But

   H-But+H2OH3O+ +ButInitial (M):0.042857M  00.042857MChange(M):x+x+xEquilibrium: 0.042857Mx+x0.042857M+x

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

  Ka=1.54×105=[H3O+][But][HBut]=x(0.042857+x)(0.042857x)=x(0.042857)0.042857[H3O+]=x=5.1333×106M

Calculation for pH and pOH

  pH=log[H3O+]=log(5.1333×106)pH=5.2896=5.29

(d)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 19.00 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa:

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(d)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 19.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 6.09.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(19.00mL)=1.900×103molNaOH

The NaOH will react with an equal amount of the aid, and 1.00×104molH-But will remain and 1.900×103 moles of But will form.

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol1.900×103mol  0Change(M)1.900×103mol1.900×103mol+1.900×103molEquilibrium  1.000×104mol01.900×103mol

Determination the liters of solution present at the equivalent point,

Volume

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

Concentration of NH4+ at equivalent point,

Molarity of the excess H-But

  =(1.00×104molH-But)/(0.03900L)=0.0025641M

Molarity of the excess Butformed is,

  =(1.900×103molBut)/(0.03900L)=0.0487179M

Using a reaction table for the equilibrium reaction of H-But

   H-But+H2OH3O+ +ButInitial (M):0.0025641M  00.0487179MChange(M):x+x+xEquilibrium: 0.0025641Mx+x0.0487179M+x

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

  Ka=1.54×105=[H3O+][But][HBut]=x(0.0487179+x)(0.0025641x)=x(0.0487179)0.0025641[H3O+]=x=8.1052632×107M

Calculation for pH and pOH

  pH=log[H3O+]=log(8.1052632×107)pH=6.09123=6.09

(e)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 19.95 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(e)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 19.95 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 6.09.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(19.95mL)=1.995×103molNaOH

The NaOH will react with an equal amount of the aid, and 5.00×104molH-But will remain and 1.995×103 moles of But will form.

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol1.995×103mol  0Change(M)1.995×103mol1.995×103mol+1.995×103molEquilibrium  5.000×106mol01.995×103mol

Determination the liters of solution present at the equivalent point,

Volume

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

Hence, volume of given solution is 0.03995L.

Molarity of the excess H-But

  =(5.00×106molH-But)/(0.03995L)=0.000125156M

Molarity of the excess Butformed is,

  =(1.995×103molBut)/(0.03995L)=0.0499374M

Using a reaction table for the equilibrium reaction of H-But

   H-But+H2OH3O+ +ButInitial (M):0.000125156M  00.0499374MChange(M):x+x+xEquilibrium: 0.000125156x+x0.0499374+x

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

  Ka=1.54×105=[H3O+][But][HBut]=x(0.0499374+x)(0.000125156x)=x(0.0499374)0.000125156[H3O+]=x=3.859637×108M

Calculation for pH and pOH

  pH=log[H3O+]=log(3.859637×108)pH=7.41354=7.41

(f)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 20.00 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(f)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 20.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 5.244.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(20.00mL)=2.000×103molNaOH

The NaOH will react with an equal amount of the aid, and 0molH-But will remain and 2.000×103 moles of But will form.

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol2.000×103mol  0Change(M)2.000×103mol2.000×103mol+2.000×103molEquilibrium  002.000×103mol

The Kb of But is now important

The volume of the solution at this point is,

Volume

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

Hence, volume of given solution is 0.04000L.

Molarity of the excess H-But

  =(5.00×106molH-But)/(0.03995L)=0.000125156M

Molarity of the excess Butformed is,

  =(2.000×103molBut)/(0.04000L)=0.05000M

So,

  Kb=KW/Ka=(1.0×1014)(1.54×105)=6.49351×1010 

Using a reaction table for the equilibrium reaction of H-But

   H-But+H2OH3O+ +ButInitial (M):0.05000M  00Change(M):x+x+xEquilibrium: 0.05000x+xx

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

  Ka=6.49351×1010=[H-But][OH][But]=[x][x](0.05000x)=x20.05000[OH]=x=5.6980304×106M

Calculation for pOH

  pOH=log[OH]=log(5.6980304×106)pOH=5.244275238

Hence, the pOH=5.244

Calculation for pH

  pH=14.00pOH=14.005.244275238=8.7557248=8.76.

(g)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 20.05 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(g)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 20.05 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 10.10.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(20.05mL)=2.005×103molNaOH

The NaOH will react with an equal amount of the aid, and 0molH-But will remain and 5.00×106 of NaOH will excess.

There will be in excess, there will be 2.000×103 mol of But produced, but this weak base will not affect the pH compared to the excess strong base NaOH.  

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol2.005×103mol  0Change(M)2.000×103mol2.000×103mol+2.000×103molEquilibrium  05.000×106mol2.000×103mol

The volume of the solution at this point is

Volume

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

Hence, volume of given solution is 0.04005L.

Molarity of the excess H-But

  =(5.00×106molH-But)/(0.04005L)=1.2484×104M

Calculation for pOH

  pOH=log[OH]=log(1.2484×104)pOH=3.9036

Hence, the pOH=3.90

Calculation for pH

  pH=14.00pOH=14.003.9036=10.0964=10.10.

(h)

Interpretation Introduction

Interpretation:

The pH during the titration 20.00 mL of 0.1000 M butanoic acid, (Ka=1.54×105 ) with 0.1000 M NaOH solution after the addition of 25.00 mL titrant has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

(h)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a given buffer solution after the addition 25.00 mL titrant and addition of 20.00 mL and 0.1000 M butanoic acid with 0.1000 M NaOH is 12.05.

Explanation of Solution

Determination the moles of NaOH added

Moles added NaOH

  =(0.1000molNaOH/L)(103L/1mL)(25.00mL)=2.500×103molNaOH

The NaOH will react with an equal amount of the aid, and 0molH-But will remain and 5.00×106 of NaOH will excess.

There will be in excess, there will be 2.000×103 mol of But produced, but this weak base will not affect the pH compared to the excess strong base NaOH.  

ICE Table of equilibrium,

 H-But(aq) +NaOH(aq)H2O(l) +But(aq)+Na+(aq)Initial (M)2.000×103mol2.500×103mol  0Change(M)2.000×103mol2.000×103mol+2.000×103molEquilibrium  05.000×104mol2.000×103mol

The volume of the solution at this point is

Volume

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

Molarity of the excess H-But

  =(5.00×106molH-But)/(0.04500L)=1.1111×102M.

Calculation for pOH

  pOH=log[OH]=log(1.1111×102)pOH=1.9542.

Hence, the pOH=1.95

Calculation for pH

  pH=14.00pOH=14.001.9542=12.0458=12.05.

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Chapter 19 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

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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Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY