The pH at the stoichiometric point of the titration of 25.0 mL of 0 .010 M HClO (aq) with 0 .020 M KOH (aq) has to be calculated. 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[H 3 O + ] On rearranging, the concentration of hydrogen ion [H + ] can be calculated using pH as follows, [H + ] = 10 -pH

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Answer 1

Question

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

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13) When solid barium phosphate is in equilibrium with its ions, the ratio of barium ions to phosphate ions would be: a. 1:1 b. 2:3 c. 3:2 d. 2:1 14) The pH of a 0.05 M solution of HCl(aq) at 25°C is 15) The pH of a 0.20 M solution of KOH at 25°C is

Answer 2

Question

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

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Answer 3

Question

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

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Answer 4

Question

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

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Answer 5

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Answer 6

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Answer 7

Chapter 6, Problem 6H.2AST

Interpretation Introduction

Interpretation:

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ has to be calculated.

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.

$pH = -log[H3O+]$

On rearranging, the concentration of hydrogen ion $[H+]$ can be calculated using pH as follows,

$[H+] = 10-pH$

Answer 8

Expert Solution & Answer

Answer to Problem 6H.2AST

The $pH$ at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The initial amount of $HClO$ present in the analyte solution is

$n(HClO) = (0.025×10-3L)×(0.01 mol×L-1) = 2.5×10-4 mol = 0.25×10-3 mol = 0.25 mmol$

The amount of $OH−$ required to react with $HClO$ is calculated and their reaction stoichiometry is

$1 mol OH- = 1 mol HClO$ .

$n(OH-) = (initial amount of HClO in the analyte)×1 mol OH-1 mol HClO = (0.25×10-3 mol)×1 mol OH-1 mol HClO = 0.25 ×10-3 mol = 0.25 mmol$

Therefore, the $OH−$ ion required to react with $HClO$ is 0.25 mmol.

From the stoichiometric point the amount of $ClO−$ is equal to the amount of $OH−$ added.

$n(ClO-) = 0.25 mmol$

The amount of $OH−$ ions present in the volume of titrant is calculated as.

$Vadded = OH-ion required to react with HClOconcentration of OH-ion = 0.25×10-3 mol0.020 mol⋅L-1 = 1.25×10-2 L = 12.5 mL$

Hence, the amount of $OH−$ ions present in the volume of titrant is 12.5 mL.

The total volume of the solution at the stoichiometric point is calculated as,

$Vfinal = Vinitial+Vadded = 25.0+12.5 mL = 37.5 mL$

The concentration of $ClO−$ ions at the stoichiometric point is

$[ClO-] = amount of ClO- present in the analyte solutiontotal volume of the solution = 0.25×10-3mol3.75×10-2 L = 0.007 mol⋅L-1$

Therefore, the concentration of $ClO−$ is $0.007 mol⋅L-1$ .

The chemical equilibrium reaction is given below.

$ClO-(aq)+H2O(l)⇌HClO(aq)+OH-(aq)$

$Kb=[HClO] [OH-][ClO-]$

	$ClO−$	$HClO$	$OH-$
Initial concentration	0.007	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.007-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

$Kb=x×x0.007-x$

The above equation, assume that the x present in 0.007-x is very small than 0.007 then it can be negligible and it follows,

$Kb=x20.007$

Hydrochlorous acid $Ka$ value is $3.0×10-8$ from this we can calculate the $Kb$ value, is given below.

$Ka×Kb=Kw Kb = KwKa = 1.0×10-143.0×10-8 = 3.33×10-7$

Therefore, the $Kb$ value of $HClO$ is $3.33×10−7$ substitute this in below equation.

$Kb=x20.0073.3×10-7= x20.007 x2 = (3.3×10-7)×0.007 x = (3.3×10-7)×0.007 = 4.81×10-5$

Hence, the concentration of $OH-$ is $4.81×10-5$ .

Now, we can calculate the pH of the solution,

$pOH =-log( 4.81×10-5) = 4.32pH+pOH = 14 pH = 14-4.32 = 9.68$

Hence, the $pH$ of the solution at the stoichiometric point of the titration of 25.0 mL of $0.010 M HClO(aq)$ with $0.020 M KOH(aq)$ is 9.68.

Answer 12

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

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