Chapter 13, Problem 98QAP

Consider the following six beakers. All have 100 mL of aqueous 0.1 M solutions of the following compounds:

beaker A has HI

beaker B has HNO₂

beaker C has NaOH

beaker D has Ba(OH)₂

beaker E has NH₄Cl

beaker F has C₂H₅NH₂

Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more Information required).

(a) The pH in beaker A the pH in beaker B.

(b) The pH in beaker C the pH in beaker D.

(c) The % ionization in beaker A the % ionization in beaker C.

(d) The pH in beaker B the pH in beaker E.

(e) The pH in beaker E the pH in beaker F.

(f) The pH in beaker C the pH in beaker F.

Interpretation Introduction

(a)

Interpretation:

The pH in beaker A and beaker B needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The pH in beaker A is less than (LT) the pH in B.

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

In beaker A, HI is a strong acid thus, the value of pH is less than the pH in beaker B which contains a weak acid ${\text{HNO}}_{\text{2}}$.

Therefore, the pH in beaker A is less than the pH in B.

Interpretation Introduction

(b)

Interpretation:

The pH in beaker C and beaker D needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The value of pH in beaker C is less than (LT) the pH in beaker D

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

The beaker C and D both contains strong base. The beaker C contains $\text{NaOH}$ and beaker D contains $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

Thus, the concentration of hydroxide ion in beaker D will be twice the concentration of hydroxide ion in beaker C.

If concentration of hydroxide ion increases, the value of pOH decreases and that of pH increases. Thus, the value of pH in beaker C is less than the pH in beaker D.

Interpretation Introduction

(c)

Interpretation:

The percent ionization in beaker A and beaker C needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The ionization percent of HF will be equal to (EQ) that of NaOH.

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

The percent ionization of a strong acid/base is more than a weak acid/base. Beaker A contains HF and beaker C contains NaOH.

Both HF and NaOH are strong and completely dissociates into their respective ions.

Therefore, ionization percent of HF will be equal to that of NaOH.

Interpretation Introduction

(d)

Interpretation:

The pH in beaker B and beaker E needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The value of pH will be less than 7 but the exact value cannot be calculated as it required more information. (MI)

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

The beaker B contains weak acid ${\text{HNO}}_{\text{2}}$ and beaker E contains ${\text{NH}}_{\text{4}}\text{Cl}$ salt.

Here, ${\text{HNO}}_{\text{2}}$ is a weak acid thus, the value of pH will be less than 7.

The salt is formed from HCl and ${\text{NH}}_{\text{4}}\text{OH}$ which is a strong acid and weak base respectively. Thus, the value of pH will be less than 7 but the exact value cannot be calculated as it required more information.

Interpretation Introduction

(e)

Interpretation:

The pH in beaker E and beaker F needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The value of pH in beaker E is less than (LT) beaker F.

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

The beaker E contains ${\text{NH}}_{\text{4}}\text{Cl}$ salt. The salt is formed from HCl and ${\text{NH}}_{\text{4}}\text{OH}$ which is a strong acid and weak base respectively. Thus, the value of pH will be less than 7.

The Beaker F contains ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base. Even if beaker F contains a weak base, the value of pH will be more than 7 in it.

Therefore, the value of pH in beaker E is less than beaker F.

Interpretation Introduction

(f)

Interpretation:

The pH in beaker C and beaker F needs to be compared.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is equilibrium concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is equilibrium concentration of conjugate base and $\left[\text{HB}\right]$ is the equilibrium concentration of weak acid.

The pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion in the solution.

Answer to Problem 98QAP

The value of pH in beaker C is greater than (GT) pH in beaker F.

Explanation of Solution

According to the question, there are 6 beakers from beaker A to E.

The aqueous solutions in beakers are as follows:

Beaker A: 0.1 M, 100 mL HI which is a strong acid.

Beaker B: 0.1 M, 100 mL ${\text{HNO}}_{\text{2}}$ which is a weak acid.

Beaker C: 0.1 M, 100 mL $\text{NaOH}$ which is a strong base.

Beaker D: 0.1 M, 100 mL $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ which is a strong base.

Beaker E: 0.1 M, 100 mL ${\text{NH}}_{\text{4}}\text{Cl}$ which is a salt.

Beaker F: 0.1 M, 100 mL ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ which is a weak base.

Beaker C contains a strong acid the pH will be highly greater than 7 and beaker F contains a weak base the value of pH will be greater than 7 but less than the value in beaker C.

Therefore, the value of pH in beaker C is greater than pH in beaker F.