Chapter 4, Problem 4.174QP

Because acid-base and precipitation reactions discussed in this chapter all involve ionic species, their progress can be monitored by measuring the electrical conductance of the solution. Match the following reactions with diagrams (a)-(d). The electrical conductance is shown in arbitrary units.

(1) A 1.0 M KOH solution is added to 1.0 L of 1.0 M CH₃COOH.

(2) A 1.0 M NaOH solution is added to 1.0 L of 1.0 M HCl.

(3) A 1.0 M BaCl₂ solution is added to 1.0 L of 1.0 M K₂SO₄.

(4) A 1.0 M NaCl solution is added to 1.0 L of 1.0 M AgNO₃.

(5) A 1.0 M CH₃COOH solution is added to 1.0 L of 1.0 M NH₃.

Interpretation Introduction

Interpretation:

The given each reaction are should be matched with given each diagram and significance of slope change points in the given diagrams should be explained.

Concept introduction:

Precipitation reaction:

• If precipitate is formed, when two solutions are mixed together is called precipitation reaction.
• The amount of precipitate formed is related to the amount of reactants taken in to the reaction.

Neutralization reaction:

• The reaction between acid and base to gives a salt is the known as neutralization reaction.

Strong and weak electrolytes:

• The compound dissolved in water and completely dissociates to produces the ions is known as strong electrolytes.
• The compound dissolved in water but not completely dissociates to produces the ions is known as strong electrolytes.

Electrical conductivity of electrolytes:

• The strong electrolytes are having high electrical conductivity than weak electrolytes.
• The number of ion in solution is directly proportional to the electrical conductivity of electrolytes.

Conductivity titration:

• The measurement of electrical conductivity of titration mixture to gives a end point if the reaction.
• The sudden change in the slope is a equivalent point of the titration and it is the end point.

To find the electrical conductance, when$\text{1}\text{.0M}\text{KOH}$ solution is added to$\text{1}\text{.0M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$

Answer to Problem 4.174QP

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.

Record the given data

Fig.1

Explanation of Solution

If the conductance unit will be twice its concentration (molarity), when compound is completely dissociates into equal number of ions in solution.

Reaction of $\text{KOH}$ with ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is,

${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2(aq)}}{\text{ + KOH}}_{\text{(aq) }}\to {\text{ KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Volume of $\text{KOH}$ is zero means Conductance unit is,

$\text{0L}KOH\text{[HC2H3O2] =}\text{0}\text{unit}$

If $\text{1L}\text{of}\text{KOH}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}KOH{\text{[KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{] = }\frac{\text{1}\text{.0 mol}}{2.0L}\\ =0.50M\\ \text{=}1\text{unit}\end{array}$

If $\text{2L}\text{of}\text{KOH}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}KOH{\text{[KC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{] = }\frac{\text{1}\text{.0 mol}}{3.0L}\\ =\frac{1}{3}M\\ \text{=}1.3\text{unit}\end{array}$

If Conductance unit of $\text{KOH}$ is,

$\begin{array}{l}\text{[KOH] = }\frac{\text{1}\text{.0 mol}}{3.0L}\\ =\frac{1}{3}M\\ \text{=}1.3\text{unit}\end{array}$

Interpretation Introduction

Interpretation:

The given each reaction are should be matched with given each diagram and significance of slope change points in the given diagrams should be explained.

Concept introduction:

Precipitation reaction:

• If precipitate is formed, when two solutions are mixed together is called precipitation reaction.
• The amount of precipitate formed is related to the amount of reactants taken in to the reaction.

Neutralization reaction:

• The reaction between acid and base to gives a salt is the known as neutralization reaction.

Strong and weak electrolytes:

• The compound dissolved in water and completely dissociates to produces the ions is known as strong electrolytes.
• The compound dissolved in water but not completely dissociates to produces the ions is known as strong electrolytes.

Electrical conductivity of electrolytes:

• The strong electrolytes are having high electrical conductivity than weak electrolytes.
• The number of ion in solution is directly proportional to the electrical conductivity of electrolytes.

Conductivity titration:

• The measurement of electrical conductivity of titration mixture to gives a end point if the reaction.
• The sudden change in the slope is a equivalent point of the titration and it is the end point.

To find the electrical conductance when,$\text{1}\text{.0 M}\text{NaOH}$ solution is added to $\text{1}\text{.0 M}\text{HCl}$

Answer to Problem 4.174QP

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.

Record the given data

Fig.1

Explanation of Solution

If the conductance unit will be twice its concentration (molarity), when compound is completely dissociates into equal number of ions in solution.

Reaction of $\text{NaOH}$ with $\text{HCl}$ is,

$HC{l}_{\text{(aq)}}{\text{ + NaOH}}_{\text{(aq) }}\to {\text{ NaCl}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Volume of $\text{NaOH}$ is zero means Conductance unit is,

$\text{0L}NaOH\text{[HCl] }\text{=}2\text{units}$

If $\text{1L}\text{of}Na\text{OH}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}NaOH\text{[NaCl] = }\frac{\text{1}\text{.0 mol}}{2.0L}\\ =0.50M\\ \text{=}1\text{unit}\end{array}$

If $\text{2L}\text{of}Na\text{OH}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}NaOH\text{[NaCl] = }\frac{\text{1}\text{.0 mol}}{3.0L}\\ =\frac{1}{3}M\\ \text{=}1.3\text{unit}\end{array}$

Interpretation Introduction

Interpretation:

The given each reaction are should be matched with given each diagram and significance of slope change points in the given diagrams should be explained.

Concept introduction:

Precipitation reaction:

• If precipitate is formed, when two solutions are mixed together is called precipitation reaction.
• The amount of precipitate formed is related to the amount of reactants taken in to the reaction.

Neutralization reaction:

• The reaction between acid and base to gives a salt is the known as neutralization reaction.

Strong and weak electrolytes:

• The compound dissolved in water and completely dissociates to produces the ions is known as strong electrolytes.
• The compound dissolved in water but not completely dissociates to produces the ions is known as strong electrolytes.

Electrical conductivity of electrolytes:

• The strong electrolytes are having high electrical conductivity than weak electrolytes.
• The number of ion in solution is directly proportional to the electrical conductivity of electrolytes.

Conductivity titration:

• The measurement of electrical conductivity of titration mixture to gives a end point if the reaction.
• The sudden change in the slope is a equivalent point of the titration and it is the end point.

To find the electrical conductance when, $\text{1}{\text{.0 M BaCl}}_{\text{2}}$ solution is added to $\text{1}{\text{.0 M K}}_{\text{2}}{\text{SO}}_{\text{4}}$

Answer to Problem 4.174QP

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.

Record the given data

Fig.1

Explanation of Solution

If the conductance unit will be twice its concentration (molarity), when compound is completely dissociates into equal number of ions in solution.

Reaction of ${\text{BaCl}}_{\text{2}}$ with ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

${\text{BaCl}}_{\text{2}}{}_{\text{(aq)}}{\text{ + K}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq) }}\to {\text{ BaSO}}_{\text{4}}{}_{\text{(s)}}{\text{ + 2KCl}}_{\text{(aq)}}$

Volume of ${\text{BaCl}}_{\text{2}}$ is zero means Conductance unit is,

$\text{0L}{\text{BaCl}}_{\text{2}}{\text{[K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{] =}4\text{units}$

If $\text{1L}{\text{BaCl}}_{\text{2}}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}{\text{BaCl}}_{\text{2}}\text{[KCl] = }\frac{\text{2}\text{.0 mol}}{2.0L}\\ =1.0M\\ \text{=}2\text{unit}\end{array}$

If $\text{2L}{\text{BaCl}}_{\text{2}}$ is added means Conductance unit is,

$\begin{array}{l}\text{2L}BaC{l}_2\text{[KCl] = }\frac{\text{2}\text{.0 mol}}{3.0L}\\ =\frac{2}{3}M\\ \text{=}2.7\text{unit}\end{array}$

Interpretation Introduction

Interpretation:

The given each reaction are should be matched with given each diagram and significance of slope change points in the given diagrams should be explained.

Concept introduction:

Precipitation reaction:

• If precipitate is formed, when two solutions are mixed together is called precipitation reaction.
• The amount of precipitate formed is related to the amount of reactants taken in to the reaction.

Neutralization reaction:

• The reaction between acid and base to gives a salt is the known as neutralization reaction.

Strong and weak electrolytes:

• The compound dissolved in water and completely dissociates to produces the ions is known as strong electrolytes.
• The compound dissolved in water but not completely dissociates to produces the ions is known as strong electrolytes.

Electrical conductivity of electrolytes:

• The strong electrolytes are having high electrical conductivity than weak electrolytes.
• The number of ion in solution is directly proportional to the electrical conductivity of electrolytes.

Conductivity titration:

• The measurement of electrical conductivity of titration mixture to gives a end point if the reaction.
• The sudden change in the slope is a equivalent point of the titration and it is the end point.

To find the electrical conductance, when$\text{1}\text{.0 M NaCl}$ solution is added to $\text{1}\text{.0 M AgNO}$

Answer to Problem 4.174QP

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.

Record the given data

Fig.1

Explanation of Solution

If the conductance unit will be twice its concentration (molarity), when compound is completely dissociates into equal number of ions in solution.

Reaction of $\text{NaCl}$ with $AgN{O}_3$ is,

$NaC{l}_{\text{(aq)}}{\text{ + AgNO}}_{\text{3(aq) }}\to {\text{ AgCl}}_{\text{(s)}}{\text{ + NaNO}}_{\text{3(aq)}}$

Volume of $\text{NaCl}$ is zero means Conductance unit is,

$\text{0L}\text{NaCl}[{\text{AgNO}}_{\text{3}}\text{] =}2\text{units}$

If $\text{1L}\text{NaCl}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}\text{NaCl}{\text{[NaNO}}_{\text{3}}\text{] = }\frac{\text{1}\text{.0 mol}}{2.0L}\\ =0.50M\\ \text{=}1\text{unit}\end{array}$

If $\text{2L}\text{NaCl}$ is added means Conductance unit is,

$\begin{array}{l}\text{2L}\text{NaCl}{\text{[NaNO}}_{\text{3}}\text{] = }\frac{\text{1}\text{.0 mol}}{3.0L}\\ =\frac{1}{3}M\\ \text{=1}\text{.3 }\text{units}\end{array}$

Interpretation Introduction

Interpretation:

The given each reaction are should be matched with given each diagram and significance of slope change points in the given diagrams should be explained.

Concept introduction:

Precipitation reaction:

• If precipitate is formed, when two solutions are mixed together is called precipitation reaction.
• The amount of precipitate formed is related to the amount of reactants taken in to the reaction.

Neutralization reaction:

• The reaction between acid and base to gives a salt is the known as neutralization reaction.

Strong and weak electrolytes:

• The compound dissolved in water and completely dissociates to produces the ions is known as strong electrolytes.
• The compound dissolved in water but not completely dissociates to produces the ions is known as strong electrolytes.

Electrical conductivity of electrolytes:

• The strong electrolytes are having high electrical conductivity than weak electrolytes.
• The number of ion in solution is directly proportional to the electrical conductivity of electrolytes.

Conductivity titration:

• The measurement of electrical conductivity of titration mixture to gives a end point if the reaction.
• The sudden change in the slope is a equivalent point of the titration and it is the end point.

To find the electrical conductance, when$\text{1}\text{.0M}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution is added to $\text{1}\text{.0M}{\text{NH}}_{\text{3}}$

Answer to Problem 4.174QP

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.

Record the given data

Fig.1

Explanation of Solution

If the conductance unit will be twice its concentration (molarity), when compound is completely dissociates into equal number of ions in solution.

Reaction of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ with ${\text{NH}}_{\text{3}}$ is,

${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2(aq)}}{\text{ + NH}}_{\text{3}}{}_{(aq)}\to N{H}_4{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$

Volume of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is zero means Conductance unit is,

$\text{0L}KOH\text{[HC2H3O2] =}\text{0}\text{unit}$

If $\text{1L}\text{of}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is added means Conductance unit is,

$\begin{array}{l}\text{1L}H{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{[}N{H}_4{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{] = }\frac{\text{1}\text{.0 mol}}{2.0L}\\ =0.50M\\ \text{=}1\text{unit}\end{array}$

If $\text{2L}\text{of}{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is added means Conductance unit is,

$\begin{array}{l}\text{2L}H{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{[}N{H}_4{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{] = }\frac{\text{1}\text{.0 mol}}{3.0L}\\ =\frac{1}{3}M\\ \text{=}0.67\text{unit}\end{array}$

Match the calculated conductance unit of each reaction in given diagrams in Fig.1.

• The reactions (2) and (4) are matched with diagram (a).
• The reaction (5) is matched with diagram (b).
• The reaction (3) is matched with diagram (c).
• The reaction (1) is matched with diagram (d).

The slope change points in the given diagrams are end or equivalent points of the tractions.