Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.149P

(a)

Interpretation Introduction

Interpretation:

[Ag+] present in the solution is (1.8×1010)[Cl-] and [AgCl2] present in the solution is (3.2×105)×[Cl-] have to be shown.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at 25oC. The concentration terms are raised to the coefficients of their respective ions. It is denoted by Ksp.

This value indicates the degree of dissociation of a compound in water. More the value of Ksp more is the solubility of the compound.

Considering an equilibrium of salt AxBy with its respective ions in solution and thus Ksp is calculated.

  AxByxAy++yBx-

  Ksp=[xAy+]x[yBx-]y

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

(a)

Expert Solution
Check Mark

Answer to Problem 19.149P

It is proved that [AgCl2]=(3.2×105)×[Cl-] and [Ag+]=(1.8×1010)[Cl-].

Explanation of Solution

Given that the equations are,

  [Ag+](aq)+[Cl-](aq)[AgCl](s)(Ksp=1.8×10-10)...(1)

  [Ag+](aq)+2[Cl-](aq)[AgCl2](aq)(Kf=1.8×105)...(2)

  Ksp=[Ag+]×[Cl-]

  [Ag+]=(1.8×1010)[Cl-] [As Ksp=1.8×10-10 ]

Subtracting the above equation (1) from equation (2),

  [AgCl](s)+[Cl-](aq)[AgCl2](aq)

  Kf=[AgCl2](aq)[Ag+](aq)×[Cl-]2(aq)...(3)

  [Ag+]=Ksp[Cl-]

Substituting the value of [Ag+] in the equation (3),

  [AgCl2]=Ksp×Kf×[Cl-]

  [AgCl2]=(1.8×1010)×(1.8×105)×[Cl-]

  [AgCl2]=(3.2×105)×[Cl-]

Therefore it is proved that [AgCl2]=(3.2×105)×[Cl-] and [Ag+]=(1.8×1010)[Cl-].

(b)

Interpretation Introduction

Interpretation:

Value of [Cl-] has to be found for the given [Ag+]=[AgCl2].

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at 25oC. The concentration terms are raised to the coefficients of their respective ions. It is denoted by Ksp.

This value indicates the degree of dissociation of a compound in water. More the value of Ksp more is the solubility of the compound.

Considering an equilibrium of salt AxBy with its respective ions in solution and thus Ksp is calculated.

  AxByxAy++yBx-

  Ksp=[xAy+]x[yBx-]y

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

(b)

Expert Solution
Check Mark

Answer to Problem 19.149P

The value of [Cl-] is equals to 2.37×10-3M

Explanation of Solution

Given that,

  [Ag+](aq)+[Cl-](aq)[AgCl](s)(Ksp=1.8×10-10)...(1)

  [Ag+](aq)+2[Cl-](aq)[AgCl2](aq)(Kf=1.8×105)...(2)

Now the condition is given that [Ag+]=[AgCl2].

  Ksp=[Ag+]×[Cl-]

  [Ag+]=(1.8×1010)[Cl-] [as Ksp=1.8×10-10 ]

From the formation constant,

  Kf=[AgCl2](aq)[Ag+](aq)×[Cl-]2(aq)...(3)

  [Ag+]=Ksp[Cl-]

Substituting the value of [Ag+] in the equation (3),

  [AgCl2]=Ksp×Kf×[Cl-]

  [AgCl2]=(1.8×1010)×(1.8×105)×[Cl-]

  [AgCl2]=(3.2×105)×[Cl-]

Now according to given condition,

  [Ag+]=[AgCl2]

  (1.8×1010)[Cl-]=(3.2×105)×[Cl-]

  [Cl-]2=(1.8×1010)(3.2×105)

   [Cl-]=2.37×10-3M

Therefore [Cl-] is equals to 2.37×10-3M.

(c)

Interpretation Introduction

Interpretation:

Plot of solubility of AgCl versus Cl- has to be explained.

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

 Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at 25oC. The concentration terms are raised to the coefficients of their respective ions. It is denoted by Ksp.

This value indicates the degree of dissociation of a compound in water. More the value of Ksp more is the solubility of the compound.

Considering an equilibrium of salt AxBy with its respective ions in solution and thus Ksp is calculated.

  AxByxAy++yBx-

  Ksp=[xAy+]x[yBx-]y

Le Chatelier’s principle:

When a system in equilibrium then is subjected to any external disturbance like change of pressure, volume, temperature etc… Then the system acts in a way to prevent that change. This is called Le-Chatelier’s principle.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given that,

  [Ag+](aq)+[Cl-](aq)[AgCl](s)(Ksp=1.8×10-10)...(1)

  Ksp=[Ag+]×[Cl-]

When the concentration of Cl- is less than the formation of AgCl is also less. Then [Ag+] is more in the solution.

As the [Cl-] starts to increase then the formation of  AgCl also increases and it continues till all the Ag+ has not been consumed. As the concentration of AgCl is increased then solubility of AgCl decreases to maintain the equilibrium. Hence equilibrium will shift to left side (According to Le Chatelier’s Principle).

After that if [Cl-] is increased then AgCl2 will form and its rate of formation increases with the increase in concentration of Cl-. As AgCl2 forms hence concentration of AgCl decreases and as a result to maintain equilibrium solubility of AgCl increases and equilibrium shifts to right.

The graph is given below,

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 19, Problem 19.149P

Figure.1

(d)

Interpretation Introduction

Interpretation:

The solubility of AgCl has to be found at the [Cl-] calculated at part (b) which is the minimum solubility of AgCl at that [Cl-].

Concept introduction:

Solubility:

Solubility is defined as the maximum amount of the solute that can be dissolved in the solvent at equilibrium.

 Solubility product constant:

Solubility product constant is defined for equilibrium between solids and its respective ions in the solution. Generally, solubility product refers only to insoluble or slightly soluble ionic substances that make equilibrium in water.

It is defined as the product of concentration of ions of a sparingly soluble salt in its saturated solution at 25oC. The concentration terms are raised to the coefficients of their respective ions. It is denoted by Ksp.

This value indicates the degree of dissociation of a compound in water. More the value of Ksp more is the solubility of the compound.

Considering equilibrium of salt AxBy with its respective ions in solution and thus the Ksp is calculated.

  AxByxAy++yBx-

  Ksp=[xAy+]x[yBx-]y

Formation constant:

A stability constant or formation constant is an equilibrium constant for the formation of a complex ion in the solution and it measures the strength of interaction between the reactants that forms the complex.

(d)

Expert Solution
Check Mark

Answer to Problem 19.149P

The solubility of AgCl at [Cl-]=2.37×10-3M is 1.52×10-7M.

Explanation of Solution

The condition given [Ag+]=[AgCl2].

From that it can be concluded that the value of [Cl-]=2.37×10-3M.

The condition is that solubility of AgCl=[Ag+]+[AgCl2].

It is already found that,

  [Ag+]=(1.8×1010)[Cl-]

  [AgCl2]=(3.2×105)×[Cl-]

  Hence, solubility of AgCl=(1.8×10-10)[Cl-]+(3.2×10-5[Cl-]=(1.8×10-10)2.37×10-3+(3.2×10-5)×2.37×10-3=1.52×10-7M

Hence the solubility of AgCl at [Cl-]=2.37×10-3M is 1.52×10-7M.

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Chapter 19 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - The scenes below depict solutions of the same...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.44PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.84PCh. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.88PCh. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.96PCh. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.100PCh. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.105PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. 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