The molar mass of acid and K a of the unknown acid has to be calculated. Concept introduction: Molar mass is the ratio of mass of given substance to that number of moles of compound. Molar mass = Amount of given substance Number of moles Acid dissociation constant K a is defined as measure of acid strength in a given solution K a = [A - ][H + ] [HA] To calculate: the molar mass of monoprotic acid

Question

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Answer 1

Question

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

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Answer 2

Question

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

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Answer 3

Question

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

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Answer 4

Question

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

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Answer 5

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

Answer 6

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

Answer 7

Chapter 17, Problem 17.22QP

(a)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the molar mass of monoprotic acid

Answer 8

(a)

Expert Solution

Answer to Problem 17.22QP

The molar mass of monoprotic acid is $1.10 × 102g/mol$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

Calculate the number of moles of monoprotic acid

$HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)$

$Number of moles of monoprotic acid = 18.4 mL × 0.0633 mol1000 mL = 0.00116 mol$

In this reaction one mole of monoprotic acid needs to neutralise one mole of $NaOH$ . The number of moles of monoprotic acid can be calculated using volume and given concentration of the formic acid.

Calculate the molar mass of monoprotic acid

$Molar mass = Amount of given substanceNumber of molesMolar mass = 0.1276 g HA0.00116 mol HA = 1.10 × 102g/mol$

The molar mass of monoprotic acid can be calculated on dividing given weight of monoprotic acid by number of moles of monoprotic acid.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The molar mass of acid and $Ka$ of the unknown acid has to be calculated.

Concept introduction:

Molar mass is the ratio of mass of given substance to that number of moles of compound.

$Molar mass = Amount of given substanceNumber of moles$

Acid dissociation constant $Ka$ is defined as measure of acid strength in a given solution

$Ka = [A-][H+][HA]$

To calculate: the number of moles of $NaOH$ and concentration of hydrogen ion

Answer 15

(b)

Expert Solution

Answer to Problem 17.22QP

The $Ka$ of the monoprotic acid is $1.6 × 10-6$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Amount of monoprotic acid is $0.1276 g$

Concentration of $NaOH$ is $0.0633 M$

$10 mL × 0.0633 mol1000 mL = 6.33 × 10-4$

$The reaction is as follows, HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Initial concentration(M): 0.0016 6.33 × 10-4 0 Change in concentration (M): 6.33 × 10-4 6.33 × 10-4 +6.33 × 10-4 Final concentration (M): 5.2 × 10-4 0 6.33 × 10-4The concentration of HA and A- are [HA] = 5.2 ×10-4 mol0.0350 L = 0.015 M [A-] = 6.33 ×10-4 mol0.0350 L = 0.0181 M pH = -log[H+] 5.87 = -log[H+] [H+] = 10-pH = 10-5.87 = 1.35 × 10-6 M$

The number of moles of sodium of hydroxide can be calculated using the given concentration divided by thousand. The weak acid reacts with $NaOH$ and the concentrations vary from the obtained concentrations, strength of hydrogen ion can be determined.

Calculate the $Ka$ of the monoprotic acid

$The equilibrium is as follows, HA(aq) → H+(aq) + A-(aq) Initial concentration(M): 0.0015 0 0.0181 Change in concentration (M): 1.35 × 10-6 +1.35 × 10-6 +1.35 × 10-6 Final concentration (M): 0.015 1.35 × 10-6 0.0181 Ka = [A-][H+][HA] = (1.35 × 10-6)(0.0181)0.015 = 1.6 × 10−6$

The $Ka$ can be calculated by using concentrations of reactants and products. By doing simple mathematical calculations the acid dissociation constant can be determined.