A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the pH of the solution at 25ºC .

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Answer 1

Textbook Question

Chapter 16, Problem 136AP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the $pH$ of the solution at $25ºC$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

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Answer 2

Textbook Question

Chapter 16, Problem 136AP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the $pH$ of the solution at $25ºC$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

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Answer 3

Textbook Question

Chapter 16, Problem 136AP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the $pH$ of the solution at $25ºC$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

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Answer 4

Textbook Question

Chapter 16, Problem 136AP

A 10.0-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500.0 mL of solution. Calculate the $pH$ of the solution at $25ºC$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$

Conclusion

The pH of the solution is calculated to be $1.18$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The pH of the solution formed by the combustion of phosphorus and consequential dissolution in water is to be determined.

Concept introduction:

The first ionization of a poly-protic acid takes place as

$H3A(aq)+H2O(l)⇌H3O+(aq)+H2A−(aq)$

$Ka1$ is the measure of the dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific to a particular temperature.

$Ka1=[H3O+][H2A−][H3A]$ …… (1)

Percent ionization is the percentage of acid that gets dissociated upon addition to water. It depends on the hydronium ion concentration.

$% dissociation=[H3O+]eq[H2A]o×100%$ …… (2)

Here, $[H3O+]eq$ is the concentration of hydronium ions at equilibrium, and $[H2A]o$ is the original acid concentration.

pH of the solution is calculated as

$pH=−log[H3O+]$ …… (3)

Molarity $(M)$ is the concentration of any substance given by number of moles $(n)$ of the substance divided by the volume of solution $(V)$ in liters.

$M=nV$ …… (4)

Answer 12

Answer to Problem 136AP

Solution:

The pH of the solution is $1.18$ .

Answer 13

Explanation of Solution

Given information:

$10.0 g$ white phosphorus is burned in excess oxygen and the product is then dissolved in enough water to make a $500.0 mL$ solution.

White phosphorous is burned in oxygen according to the reaction:

$P4(s)+5O2(g)→P4O10(s)$

According to the stoichiometry, $124 g$ of phosphorous reacts to produce $284 g$ of $P4O10$ .

So, $10.0 g$ of phosphorous will produce.

$m(P4O10)=284124×10.0=22.9 g$

Molar mass of $P4O10$ is $284 g/mol$ .

Number of moles of $P4O10$ is calculated as

$n(P4O10)=m(P4O10)MM(P4O10)=22.9 g284 g/mol=0.081 mol$

Now, $P4O10$ is dissolved in water to form $500.0 mL$ phosphoric acid solution, according to the reaction:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$

According to the stoichiometry, 1 mole of $P4O10$ produces 4 moles of $H3PO4$ . So, $0.081 mol P4O10$ will give:

$n(H3PO4)=41×0.081=0.323 mol$

Use equation (4) to calculate molarity of the solution as

$M=0.323 mol0.5 L=0.646 mol$

Refer to table $16.8$ for $Ka1, Ka2, and Ka3$ values of phosphoric acid as

$Ka1=7.5×10−3Ka2=6.2×10−8Ka3=4.8×10−13$

When phosphoric acid is dissolved in water, dissociation takes place in three steps as it is a triprotic acid. First, one proton is partially dissociated, becausephosphoric acid is a weak acid. Thus, the pH of the solution is determined by the contribution made by all the three proton dissociation steps.

The reaction of the first proton dissociation of phosphoric acid is represented as

$H3PO4(aq)+H2O(l)⇌H3O+(aq)+H2PO4−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as

$H3PO4(aq)H2O(l)H3O+(aq)H2PO4−(aq)Initial concentration(M)0.646−00Change in concentration(M)−x−+x+xEquilibrium concentration(M)0.646−x−xx$

Now, substitute these concentrations in equation (1) as

$Ka1=(x)(x)(0.646−x)$

Since the value of $Ka1$ is small, the amount of acid dissociated is less. Therefore, $(0.646−x)$ can be approximated as $0.646$ . Now, substitute the value of $Ka1$ in above equation as

$7.5×10−3=(x)(x)0.646x2=(7.5×10−3)0.646x=4.84×10−3x=0.067$

Thus, $[H3O+]1(eq)=0.067M$ , $[H3PO4](O)=0.646 M$

Calculate the percent dissociation from equation (2) as

$% ionization=0.0670.646×100%=10.77 %$

Since the percent dissociation is more than $5%$ , the approximation is not valid. Thus, again solve for the value of $x$ as

$7.5×10−3=(x)(x)(0.646−x)x2+(7.5×10−3)x=(4.84×10−3)x=−0.059, 0.066$

Since concentration cannot be negative, so $x=0.066$ .

Thus,

$[H2PO4−]=0.066 M[H3O+]=0.066 M$

The second and third ionization constants for phosphoric acid have a low value. Thus, it will have negligible contribution in the hydronium ion concentration.

Now, use equation (3) to calculate pH of the solution as

$pH=−log[0.066]=1.18$