The p H curve for the titration of 25.0 c m 3 of 0.15 M B a ( O H ) 2 w i t h 0.22 M H C l has to be drawn. Concept Introduction: For the titration of a weak base with a strong acid, the p H before the equivalence point depends on the excess concentration of base and the p H after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7 .0 . The equivalence point occurs when total volume of base has been added. p H = − log [ H 3 O + ] p H = − log [ O H − ] p H + p O H = 14 Calculation of Moles: 1. N o . o f m o l e s ( n ) = W M . W Where, W = W e i g h t o f a g i v e n c o m p o u n d M . W = M o l e c u l a r w e i g h t o f t h e s a m e c o m p o u n d 2. N o . o f m o l e s ( n ) = ( M o l a r i t y ) ( V o l u m e ) Molarity: Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation, M o l a r i t y = moles of solute volume of solution in L

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Answer 1

Question

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

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Answer 2

Question

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

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Answer 3

Question

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

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Answer 5

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

Answer 6

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3

Excess[H3O+] = 0.65×10−3 moles35.00×10−3L = 0.0185 M.

pH=−log[H3O+] = −log(0.0185) = 1.7311.

Therefore, the pH of the solution has been calculated to be 1.7311.

Now we can plot the graph as shown in figure 1.

Volume of added acid/mLpH013.1761512.71278.5227.0000101.7311

Elements Of Physical Chemistry, Chapter 5, Problem 5.37P

Answer 7

Chapter 5, Problem 5.37P

Interpretation Introduction

Interpretation:

The pH curve for the titration of 25.0 cm3 of 0.15 M Ba(OH)2 with 0.22M HCl has to be drawn.

Concept Introduction:

For the titration of a weak base with a strong acid, the pH before the equivalence point depends on the excess concentration of base and the pH after the equivalence point depends on the excess concentration of acid. At the equivalence point, there is not an excess of either acid or base so pH is 7.0. The equivalence point occurs when total volume of base has been added.

pH=−log[H3O+]

pH = −log[OH−]

pH+pOH=14

Calculation of Moles:

1. No. of moles (n)=WM.W

Where,

W = Weight of a given compoundM.W = Molecular weight of the same compound

2. No. of moles (n) = (Molarity)( Volume)

Molarity:

Molarity or molar concentration is the number of moles of solute dissolved per liter of solution, which can be calculated using the following equation,

Molarity = moles of solutevolume of solution in L

Answer 8

Expert Solution & Answer

Explanation of Solution

The reaction occurring in the titration is the neutralization of OH− (from Ba(OH)2) by H3O+ (from HCl):

Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)H3O+(aq) + OH−(aq) → 2H2O(l)

Which shows that 2 equivalent of acid react with 1 equivalent of Barium oxide. For achieving equivalence point, we need the equivalent quantities of acid and base. So

M1V1 = M2V2

Where,

M1= Concentration of base

M2= Concentration of acid

V1= Volume of base

V2= Volume of acid

Substituting the given values in the above equation and solving for V2,

M1V1 = M2V2(0.15M)(25 mL)= (2×0.22)(V2)V2= (0.15M)(25 mL)2×0.22= 8.522 mL.

For neutralization we need 8.522 mL of acid.

For plotting the graph we can start the titration by taking 0.00mL of acid and gradually add acid in dropwise manner till 8.5 mL.

Case – 1:

Given data:

Volume of Ba(OH)2 = 25.00 ml = 25.00×10−3LConcentration of Ba(OH)2 = 0.15MConcentration of HCl = 0.22MVolume of HCl added = 0.00 ml = 0.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 0.00×10−3L = 25.00×10−3L

The initial number of moles of Ba(OH)2= Molarity×Volume = (0.15 M)(25.00×10−3L) = 3.75×10−3.

Moles of added HCl = Molarity×Volume = (0.22M)(0.00×10−3L) =0.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 0 − 0Change 0 0 − 0_Final 3.75×10−3 0 − 0

Excess[OH−] = 3.75×10−3 moles25.00×10−3L = 0.15 M.

pOH=−log[OH−] = −log(0.15) = 0.8239.

pH = 14 − pOH = 14−0.8239=13.1761.

Therefore, the initial pH of the solution has been calculated to be 13.1761.

Case – 2:

Volume of HCl added = 5.00 ml = 5.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 5.00×10−3L = 30.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(5.00×10−3L) = 1.1×10−3.

As there is two molecules of HCl, moles of added HCl will be 2.2×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 2.2×10−3 − 0Change −2.2×10−3 − 2.2×10−3 − +2.2×10−3_Final 1.55×10−3 0 − 2.2×10−3

Excess[OH−] = 1.55×10−3 moles30.00×10−3L = 0.0516M.

pOH=−log[OH−] = −log(0.0561) = 1.2873.

pH = 14 − pOH = 14−1.2873=12.7127.

Therefore, the pH of the solution has been calculated to be 12.7127.

Case – 3:

Volume of HCl added = 8.522 ml = 8.522×10−3LThe total volume of the solution at this point = 25.00×10−3L + 8.522×10−3L = 33.522×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(8.522×10−3L) = 1.8748×10−3.

As there is two molecules of HCl, moles of added HCl will be 3.75×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 3.75×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0 − 3.75×10−3

The HCl will react with an equal amount of the Ba(OH)2 and 0.0 mol Ba(OH)2 will remain. This is the equivalence point of a weak base –strong acid titration, thus, the pH is 7.0. Only the neutral salt BaCl2 is in solution at the equivalence point.

Case – 4:

Volume of HCl added = 10.00 ml = 10.00×10−3LThe total volume of the solution at this point = 25.00×10−3L + 10.00×10−3L = 35.00×10−3L

Moles of added HCl = Molarity×Volume = (0.22M)(10.00×10−3L) = 2.2×10−3.

As there is two molecules of HCl, moles of added HCl will be 4.4×10−3.

No. of moles Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)_ Initial 3.75×10−3 4.4×10−3 − 0Change −3.75×10−3 − 3.75×10−3 − +3.75×10−3_Final 0 0.65×10−3 − 3.75×10−3