“The neutral solution has a p H o f 7 ”should be explained Concept Introduction: ► p H of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution. Thus, p H = − log [ H + ] ► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions. H 2 O ( l ) + H 2 O ( l ) ↔ H + ( a q ) + O H − ( a q ) Applying law of mass action, the equilibrium constant K c for the self ionization of water can be written as K c = [ H + ] [ O H − ] [ H 2 O ] 2 K c [ H 2 O ] 2 = [ H + ] [ O H − ] Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore [ H 2 O ] 2 may be regarded as constant. K c K = [ H + ] [ O H − ] where K is a constant representing [ H 2 O ] 2 Now, K w = [ H + ] [ O H − ] Where K w is the product of two constant K c a n d K K w is called the ionic product of water and its value is 1 × 1 0 − 14

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Answer 1

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Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

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Answer 2

Question

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

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Answer 3

Question

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

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Answer 4

Question

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

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Check out a sample textbook solution

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Answer 5

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

Answer 6

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

Answer 7

Chapter 11.6, Problem 11.43QAP

Interpretation Introduction

Interpretation: “The neutral solution has a $p H o f 7$ ”should be explained

Concept Introduction:

► $p H$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

Thus, $p H = - \log [H^{+}]$

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_{2} O_{(l)} + H_{2} O_{(l)} \leftrightarrow H^{+}_{(a q)} + O H^{-}_{(a q)}$

Applying law of mass action, the equilibrium constant $K_{c}$ for the self ionization of water can be written as

$\begin{array}{l} K_{c} = \frac{[H^{+}] [O H^{-}]}{{[H_{2} O]}^{2}} \\ K_{c} {[H_{2} O]}^{2} = [H^{+}] [O H^{-}] \end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${[H_{2} O]}^{2}$ may be regarded as constant.

$K_{c} K = [H^{+}] [O H^{-}]$

where $K$ is a constant representing ${[H_{2} O]}^{2}$

Now,

$K_{w} = [H^{+}] [O H^{-}]$

Where $K_{w}$ is the product of two constant $K_{c} a n d K$

$K_{w}$ is called the ionic product of water and its value is $1 \times 1 0^{- 14}$

Answer 8

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Answer 9

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Answer 10

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Answer 11

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