The curve for the titration of 50 m L o f a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of N a O H has to be sketched.

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Answer 1

Question

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

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Answer 2

Question

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

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Answer 3

Question

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

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Answer 4

Question

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

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Answer 5

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

Answer 6

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

Answer 7

Chapter 16, Problem

Interpretation Introduction

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Answer 8

Expert Solution & Answer

Explanation of Solution

To sketch a titration curve for a weak base with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The initial point in the titration when the acid is not added to the system, the system is a weak base system.

The iCe table can be set up to calculate the pH of the solution.

B+H2O⇌OH−+HBi(M)0.100C(M)−y+y+ye(M)0.1−yyy

Where, B represents weak base and HB represents the conjugate acid of weak base.

The Kb value for weak base is 1.0×10−5. The expression for Kb can be written as given below,

Kb=[OH−][HB][B]=1.0×10−5y2(0.1−y)=1.0×10−5

This can be solved by approximation. If y≪0.10, then

y2≈1.0×10−5×0.10=1.0×10−6y=1.0×10−6=1.0×10−3=[OH−]

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(1.0×10−3)=3.00pH=14−pOH=14−3.00=11.00.

Addition of 10 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

The neutralization reaction can be written as given,

B(aq)+H3O+(aq)→HB(aq)+H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimol

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

B+H3O+→HB+H2Os(mmol)520ExcessR(mmol)−2−2+2+2f(mmol)302Excess

All the strong acid is consumed. There are weak base (B) and its conjugate acid (HB) in the solution. So this solution is a basic buffer. By using Henderson-Hasselbalch equation, it’s pH can be calculated.

pOH=pKb+lognanb=pKb+log2.03.0=−log(Kb)−0.176=−log(1.0×10−5)−0.176=5.00−0.176=4.824pH=14−pOH=14−4.824=9.176.

Addition of 25 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=25.00+50.00=75.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

B+H3O+→HB+H2Os(mmol)550ExcessR(mmol)−5−5+5+5f(mmol)005Excessc(M)000.067

Both H3O+ and weak base have been completely consumed, the titration is at the equivalence point. There is only the conjugate acid and water in the solution.

The iCe table can be set up to calculate the pH of the solution.

HB+H2O⇌B+H3O+i(M)0.06700C(M)−y+y+ye(M)0.067−yyy

The expression for Ka can be written as given below,

Ka=[B][H3O+][HB]=y2(0.067−y)

The value of Ka can be calculated as given below.

Ka=KwKb=1.0×10−141.0×10−5=1.0×10−9

This can be solved by approximation. If y≪0.067, then

y2≈1.0×10−9×0.067=0.067×10−9y=0.067×10−9=8.2×10−6=[H3O+]

The pH of the solution can be calculated as given below.

pH=−log[H3O+]=−log(8.2×10−6)=5.1.

Addition of 50 mL of 0.20 M HClto50.00 mL of 0.10 Mweak base:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of B=50.00 mL×(0.10 milimol mL)=5 milimolTotal volume=50.00+50.00=100.00 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is the base.

B+H3O+→HB+H2Os(mmol)5100ExcessR(mmol)−5−5+5+5f(mmol)055Excessc(M)00.050.05

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

To sketch a titration curve for sodium hydroxide with a strong acid, the given below example can be taken where HCl has been taken as strong acid.

Addition of 0 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The initial point in the titration when the acid is not added to the system, the system is a strong base system.

As NaOH is a strong base, the hydroxide ion concentration is 0.10 M.

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.1)=1pH=14−pOH=14−1=13.

Addition of 10 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

The neutralization reaction can be written as given,

H3O+(aq)+OH−(aq)→2H2O(l)

Calculation of milimoles of acid and base:

Amount of HCl=10.00 mL×(0.20 milimol H3O+mL)=2 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=60 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is H3O+ ion.

H3O++OH−→2H2Os(mmol)25ExcessR(mmol)−2−2+2f(mmol)03ExcessC(M)00.05

The pH of the solution can be calculated as given below.

pOH=−log[OH−]=−log(0.05)=1.30pH=14−pOH=14−1.30=12.7.

Addition of 25 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base:

Amount of HCl=25.00 mL×(0.20 milimol H3O+mL)=5 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=75 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1.

H3O++OH−→2H2Os(mmol)55ExcessR(mmol)−5−5+5f(mmol)00ExcessC(M)00Excess

Both H3O+ and OH− ions have been completely consumed, the titration is at the equivalence point. Therefore, the pH of the system is 7.

Addition of 50 mL of 0.20 M HCl to50.00 mL of 0.10 M NaOH:

Calculation of milimoles of acid and base and total volume:

Amount of HCl=50.00 mL×(0.20 milimol H3O+mL)=10 milimolAmount of OH−=50.00 mL×(0.10 milimol OH−mL)=5 milimolTotal volume=100 mL

Making of sRfc table:

The table can be formed as shown below.

The molar ratio of acid and base is 1:1. So the limiting reactant is OH− ion.

H3O++OH−→2H2Os(mmol)105ExcessR(mmol)−5−5+5f(mmol)50ExcessC(M)0.050Excess

Now, the pH of the system can be calculated as given below.

pH=−log[H3O+]=−log(0.05)=1.30.

Titration curve:

The titration curve is plotted between volume of acid added and the corresponding pH values.

For weak base:

Volume of HCl(mL)pH011.0010.009.17625.005.1050.001.30

For strong base:

Volume of NaOH(mL)pH013.0010.0012.7025.007.0050.001.30

The titration curve with four important regions is given below.

Chemistry Principles And Practice, Chapter 16, Problem 16.57QE

Figure

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 16 - Prob. 16.1QE Ch. 16 - Sketch a titration curve for the titration of...Ch. 16 - Prob. 16.4QE Ch. 16 - Prob. 16.5QE Ch. 16 - Explain why the HendersonHasselbalch equation...Ch. 16 - Prob. 16.7QE Ch. 16 - Prob. 16.8QE Ch. 16 - Prob. 16.9QE Ch. 16 - Prob. 16.11QE Ch. 16 - Prob. 16.13QE

The curve for the titration of 50 m L o f a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of N a O H has to be sketched.

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Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Explanation of Solution

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Chapter 16 Solutions

The curve for the titration of 50 m L o f a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of N a O H has to be sketched.

Concept explainers

Videos

Interpretation: The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.

Explanation of Solution

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Chapter 16 Solutions

Interpretation:

The curve for the titration of 50 mL of a 0.10 M weak base with a 0.20 M strong acid has to be sketched. On the same axes, the titration curve for the same volume and concentration of NaOH has to be sketched.