The pH of 0.00115 M HCl and 0.0100 M HClO 3 solution needed to be calculated. Concept introduction: The of any solution is defined as the negative logarithm of the hydrogen ion concentration. p H = − log [ H + ] The dissociation reaction of the HA weak acid, with H + (aq) as product is as follows: H A ( a q ) → H + ( a q ) + A − ( a q ) The expression for ionization constant ( K a )is as follows: K a = [ A − ] [ H + ] [ H A ] The auto-ionization of water can be written as follows: H 2 O ( l ) → H + ( a q ) + O H − ( a q ) Therefore; K w = [ H 3 O + ] [ O H − ] = [ H + ] [ O H − ] = 1.0 × 10 − 14

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Answer 1

Question

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

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Answer 2

Question

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

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Answer 3

Question

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

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Answer 4

Question

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

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Answer 5

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

Answer 6

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

Answer 7

Chapter 17, Problem 147E

Interpretation Introduction

Interpretation:

The pHof 0.00115 M HCl and 0.0100M HClO₃ solution needed to be calculated.

Concept introduction:

     The of any solution is defined as the negative logarithm of the hydrogen ion concentration.

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

     The dissociation reaction of the HA weak acid, with H⁺(aq) as product is as follows:

$H A (a q) \to H^{+} (a q) + A_{}^{-} (a q)$

     The expression for ionization constant (K_a)is as follows:

$K_{a} = \frac{[A_{}^{-}] [H^{+}]}{[H A]}$

     The auto-ionization of water can be written as follows:

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

     Therefore;

$K_{w} = [H_{3} O^{+}] [O H^{-}] = [H^{+}] [O H^{-}] = 1.0 \times 10^{- 14}$

Answer 8

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Answer 9

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Answer 10

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Answer 11

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Solution Summary: The author explains the dissociation equation of 0.0100 M HClO 2 and the expression for ionization constant.

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