The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of 25 ° C is to becalculated. Concept introduction: Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows: K a = [ products ] [ reactants] . Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows: Molarity = Moles of solute Liter of solution .

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Answer 1

Question

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

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Students have asked these similar questions

A small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…

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Answer 2

Question

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

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Answer 3

Question

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

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Answer 4

Question

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

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Answer 5

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

Answer 6

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows:

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

Rearrange the above equation:

$(x2+7.5×10−3x−4.85×10−3)=0$ .

This is a quadratic equation, by solving:

$x=0.066 M.Therefore, [H3O+]=0.066 M.$

The pH of the solution is to be calculated as follows:

$pH=−log[H+]$ .

Here, the concentration of hydronium ions $([H3O+])$ is equal to the concentration of $[H+]$ .

$pH =−log(0.066)pH=1.18.$

Conclusion

The pH of the solution at $25°C$ is $1.18$ .

Answer 7

Chapter 24, Problem 101AP

Interpretation Introduction

Interpretation:

The pH of the solution, obtained by dissolving the product, of the given reaction, in water, at a temperature of $25°C$ is to becalculated.

Concept introduction:

Equilibrium constant isdefined as the ratio of concentration of products to the concentration of reactants and is expressed as follows:

$Ka=[products][reactants]$ .

Molarity is defined as the number of moles of solute dissolved in one litre of solution and is expressed as follows:

$Molarity=Moles of soluteLiter of solution$ .

Answer 8

Expert Solution & Answer

Answer to Problem 101AP

Solution: pH is $5.72$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information: The white phosphorous burnt is $10.0 g$ .

The volume of solution is $0.500 L$ .

If the white phosphorus burns in excess of oxygen, the reaction is as follows:

$P4(s)+5O2(g)→P4O10(s)$

The product formed $(P4O10)$ is dissolved in water and the reaction that takes place is as follows:

$P4O10(s)+6H2O(l)→4H3PO4(aq)$ .

The moles formed by $H3PO4$ are to be calculated as follows:

$Moles=Given massMolar mass$ .

Substitute the values in the above equation:

$MH3PO4=10.0 g P4×(1 mol P4123.9 g P4)(1 mol P4O101 mol P4)(4 mol H3Po41 mol P4O10)MH3PO4=0.323 mol H3Po4.$

Thus, the moles of $H3PO4$ are $0.323 mol$ .

Molarity of the phosphoric acid solution is calculated as follows:

$Molarity=Moles of soluteVolume of Solution (L)$ .

Substitute $0.323 mol$ for the moles of solute and $0.500 L$ for the volume of solution in the above equation as follows:

$Molarity=(0.323 mol0.500 L);Molarity=0.646 M.$

The equilibrium expression for a phosphoric acid solution is as follows:

$H3PO4(aq) + H2O(aq)⇌H3O+(aq) + H2PO−(aq)initial(M): 0.646 0 0change(M): -x +x +x_Equilibrium(M): 0.646 -x x x$

The value of $Ka$ for phosphoric acid is $7.5×10−3$ , taken from the reference table $16.8$ .

The $[H3O+]$ is calculated with the help of the equilibrium constant as follows:

$Ka=[H3O+][H2PO4−][H3PO4]$ .

Substitute $(0.646−x)$ for the concentration of phosphoric acid, $(x)$ for the concentration of hydronium ions, $(x)$ for the concentration of dihydrogen phosphate ions, and $7.5×10−3$ for $Ka$ in the above equation as follows: