The p H of the solution that results from mixing 30.0 m L o f 0.050 M H C N ( a q ) a n d 70.0 m L o f 0.030 M N a C N ( a q ) has to be calculated. Concept Introduction: Henderson – Hasselbalch equation: The p H of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below. p H = p K a + log [ b a s e ] i n i t i a l [ a c i d ] i n i t i a l = p K a + log ( m o l e s o f b a s e m o l e s o f a c i d )

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Answer 1

Question

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

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Answer 2

Question

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

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Answer 3

Question

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

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Answer 4

Question

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

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Answer 5

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 6

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

Answer 7

Chapter 6, Problem 6G.9E

(a)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 30.0 mL of 0.050 M HCN(aq) and 70.0 mL of 0.030 MNaCN(aq) has to be calculated.

Concept Introduction:

Henderson – Hasselbalch equation:

The pH of a buffer can be easily calculated by using Henderson – Hasselbalch equation as given below.

pH=pKa+log[base]initial[acid]initial=pKa+log(moles of basemoles of acid)

Answer 8

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 9

(b)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 10

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 11

(c)

Interpretation Introduction

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

Interpretation: The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated. Concept Introduction: Refer to part (a).

Interpretation: The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated. Concept Introduction: Refer to part (a).

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Chapter 6 Solutions

Interpretation:

The pH of the solution that results from mixing 40.0 mL of 0.030 M HCN(aq) and 60.0 mL of 0.050 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).

Interpretation:

The pH of the solution that results from mixing 25.0 mL of 0.105 M HCN(aq) and 25.0 mL of 0.105 MNaCN(aq) has to be calculated.

Concept Introduction:

Refer to part (a).