The formula for the conjugate acid of NH 3 has to be given. Concept introduction: Conjugate acid-Base pairs: When an acid is dissolved in water, the acid (HA) donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base). RCOOH has one more H + than RCOO – , similarly H 3 O + has one more H + than H 2 O therefore, RCOOH and H 3 O + are acids, RCOO – and H 2 O bases. The conjugate acid/base pairs are RCOOH / RCOO – and H 3 O + / H 2 O .

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Answer 1

Question

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

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Answer 2

Question

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

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Answer 3

Question

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

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Answer 4

Question

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

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Answer 5

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 6

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 7

Chapter 18, Problem 18.14P

(a)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH3$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 8

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 9

(b)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $NH2−$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 10

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 11

(c)

Interpretation Introduction

Interpretation:

The formula for the conjugate acid of $C10H14N2$ has to be given.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $(HA)$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$RCOOH$ has one more $H+$ than $RCOO–$ , similarly $H3O+$ has one more $H+$ than $H2O$ therefore, $RCOOH$ and $H3O+$ are acids, $RCOO–$ and $H2O$ bases.

The conjugate acid/base pairs are $RCOOH / RCOO– and H3O+/ H2O$ .

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 18.2 - Prob. 18.1AFP Ch. 18.2 - Prob. 18.1BFP Ch. 18.2 - Prob. 18.2AFP Ch. 18.2 - Prob. 18.2BFP Ch. 18.2 - Prob. 18.3AFP Ch. 18.2 - Prob. 18.3BFP Ch. 18.3 - Prob. 18.4AFP Ch. 18.3 - Prob. 18.4BFP Ch. 18.4 - Prob. 18.5AFP Ch. 18.4 - Prob. 18.5BFP

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