Chapter 16, Problem 16.157QP

Weak base B has a pK_b of 6.78 and weak acid HA has a pK_a of 5.12.

1. a Which is the stronger base, B or A⁻?
2. b Which is the stronger acid, HA or BH⁺?
3. c Consider the following reaction:

$\text{B}(aq)+\text{HA}(aq)\stackrel{}{\rightleftharpoons }{\text{BH}}^{+}(aq)+{\text{A}}^{-}(aq)$

Based on the information about the acid/base strengths for the species in this reaction, is this reaction favored to proceed more to the right or more to the left? Why?

1. d An aqueous solution is made in which the concentration of weak base B is one half the concentration of its acidic salt, BHCl, where BH⁺ is the conjugate weak add of B. Calculate the pH of the solution.
2. e An aqueous solution is made in which the concentration of weak acid HA twice the concentration of the sodium salt of the weak acid, NaA. Calculate the pH of the solution.
3. f Assume the conjugate pairs B/BH⁺ and HA/A⁻ are capable of being used as color-based end point indicators in acid–base titrations, where B is the base form indicator and BH⁻ is the acid form indicator, and HA is the acid form indicator and A⁻ is the base form indicator. Select the indicator pair that would be best to use in each of the following titrations:
   1. (1) Titration of a strong acid with a strong base.
      1. (i) B/BH⁺
      2. (ii) HA/A⁻
   2. (2) Titration of a weak base with a strong acid.
      1. (i) B/BH⁺
      2. (ii) HA/A⁻

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

The stronger base, $\text{B}$ or ${\text{A}}^{-}$ has to be found

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The stronger base is $\text{B}$

Explanation of Solution

To Find: The stronger base, $\text{B}$ or ${\text{A}}^{-}$

Given data:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12

The stronger base, $\text{B}$or ${\text{A}}^{-}$:

The strong or weak base can be found from the ${\text{pK}}_b$ value.

The given ${\text{pK}}_b$ value for base is 6.78

The given ${\text{pK}}_a$ value for acid is 5.12

The ${\text{pK}}_b$ value for ${\text{A}}^{-}$ is calculated as follows and compared.

$\begin{array}{l}{\text{pK}}_{\text{a}}{\text{ + pK}}_{\text{b}}\text{=14}\\ {\text{ pK}}_{\text{b}}=14-{\text{pK}}_{\text{a}}\\ =14-5.12\\ =8.88\end{array}$

The strong base will have small ${\text{pK}}_b$ value

The ${\text{pK}}_b$ value of ${\text{A}}^{-}$ is greater than $\text{B}$

Hence, $\text{B}$ is the stronger base

Conclusion

The stronger base was found as $\text{B}$

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

The stronger acid, $\text{HA}$ or ${\text{BH}}^{+}$ has to be found

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The stronger acid is $\text{HA}$

Explanation of Solution

To Find: The stronger acid, $\text{HA}$ or ${\text{BH}}^{+}$

The stronger acid, $\text{HA}$ or ${\text{BH}}^{+}$:

The strong or weak acid can be found from the ${\text{pK}}_a$ value.

The given ${\text{pK}}_b$ value for base is 6.78

The given ${\text{pK}}_a$ value for acid is 5.12

The ${\text{pK}}_a$ value for ${\text{BH}}^{+}$ is calculated as follows and compared.

$\begin{array}{l}{\text{pK}}_{\text{a}}{\text{ + pK}}_{\text{b}}\text{=14}\\ {\text{ pK}}_a=14-{\text{pK}}_{\text{a}}\\ =14-6.78\\ =7.22\end{array}$

The strong acid will have small ${\text{pK}}_a$ value

The ${\text{pK}}_a$ value of ${\text{BH}}^{+}$ is greater than $\text{HA}$

Hence, $\text{HA}$ is the stronger acid

Conclusion

The stronger acid was found as $\text{HA}$

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

Consider the reaction ${\text{B}}_{\text{(aq)}}{\text{ + HA}}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{BH}}^{+}{}_{(\text{aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$, whether the given reaction favored to proceed more to the right or more to the left has to be explained

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The given reaction favored to proceed more to the right side

Explanation of Solution

To Explain: Whether the given reaction favored to proceed more to the right or more to the left

Given data:

Consider the reaction ${\text{B}}_{\text{(aq)}}{\text{ + HA}}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{BH}}^{+}{}_{(\text{aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid-base neutralization reaction favours the side of reaction bearing weaker base and weaker acid.

Hence, the given reaction will be favoured more to the right side

Conclusion

The given reaction favored to proceed more to the right side

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

The pH of the aqueous solution in which the concentration of weak base $\text{B}$ is one half the concentration on its acidic salt, $\text{BHCl}$ has to be calculated

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The pH of the aqueous solution in which the concentration of weak base $\text{B}$ is one half the concentration on its acidic salt, $\text{BHCl}$ was calculated is 6.92

Explanation of Solution

To Calculate: The pH of the aqueous solution in which the concentration of weak base $\text{B}$ is one half the concentration on its acidic salt, $\text{BHCl}$

Given data:

An aqueous solution is prepared in which the concentration of the weak base $\text{B}$ is one half the concentration on its acidic salt, $\text{BHCl}$

pH calculation:

$\text{BHCl}$ salt contains ${\text{BH}}^{+}$ and ${\text{Cl}}^{-}$ ions.

The prepared solution is a buffer solution composing the weak base $\text{B}$ and its conjugate acid ${\text{BH}}^{+}$

Using Henderson-Hasselbalch equation, the pH can be calculated as follows,

${\text{pH = pK}}_a+\text{ log}\frac{\text{[base]}}{\text{[acid]}}$

The ${\text{pK}}_a$ value for ${\text{BH}}^{+}$ is 7.22

On considering the concentration of $\text{BHCl}$ as x M, then the concentration of base $\text{B}$ will be 0.5x M. Thus,

$\begin{array}{l}\text{pH }\text{= 7}\text{.22}+\text{ log}\frac{(0.5x)}{x}\\ =\text{7}\text{.22}+\text{ log(0}\text{.50)}\\ =6.92\end{array}$

Therefore, the pH of the given solution is 6.92

Conclusion

The pH of the aqueous solution in which the concentration of weak base $\text{B}$ is one half the concentration on its acidic salt, $\text{BHCl}$ was calculated as 6.92

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

The pH of the aqueous solution in which the concentration of weak acid $\text{HA}$ is twice the concentration of the sodium salt of the weak acid, $\text{NaA}$ has to be calculated.

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The pH of the aqueous solution in which the concentration of weak acid $\text{HA}$ is twice the concentration of the sodium salt of the weak acid, $\text{NaA}$ is4.82

Explanation of Solution

To Calculate: The pH of the aqueous solution in which the concentration of weak acid $\text{HA}$ is twice the concentration of the sodium salt of the weak acid, $\text{NaA}$

Given data:

An aqueous solution is prepared in which the concentration of the weak acid $\text{HA}$ is twice the concentration of the sodium salt of the weak acid, $\text{NaA}$

pH calculation:

$\text{NaA}$ salt contains ${\text{Na}}^{+}$ and ${\text{A}}^{-}$ ions.

The prepared solution is a buffer solution composing the weak acid $\text{HA}$ and its conjugate base ${\text{A}}^{-}$

Using Henderson-Hasselbalch equation, the pH can be calculated as follows,

${\text{pH = pK}}_a+\text{ log}\frac{\text{[base]}}{\text{[acid]}}$

The ${\text{pK}}_a$ value for ${\text{A}}^{-}$ is 5.12

On considering the concentration of $\text{HA}$ as x M, then the concentration of base ${\text{A}}^{-}$ will be 0.5x M. Thus,

$\begin{array}{l}\text{pH }\text{= 5}\text{.12}+\text{ log}\frac{(0.5x)}{x}\\ =5.12+\text{ log(0}\text{.50)}\\ =4.82\end{array}$

Therefore, the pH of the given solution is 4.82

Conclusion

The pH of the aqueous solution in which the concentration of weak acid $\text{HA}$ is twice the concentration of the sodium salt of the weak acid, $\text{NaA}$ was calculated as 4.82

Interpretation Introduction

Interpretation:

A weak base $\text{B}$ has a ${\text{pK}}_b$ of 6.78 and a weak acid $\text{HA}$ has a ${\text{pK}}_a$ of 5.12.

The conjugate pairs that can be used as a best indicator for the given titrations has to be selected.

1. (1) Titration of strong acid with a strong base
2. (2) Titration of weak base with a strong acid

Concept Introduction:

Strength of acid and base:

The strength of acid and base depends on the ${\text{pK}}_a$ and ${\text{pK}}_b$ value.

• Smaller the ${\text{pK}}_a$ value, stronger is the acid
• Larger the ${\text{pK}}_a$ value, weaker is the acid
• Smaller the ${\text{pK}}_b$ value, stronger is the base
• Larger the ${\text{pK}}_b$ value, weaker is the base

Henderson-Hasselbalch equation :

The pH can be calculated using Henderson-Hasselbalch equation as follows,

$\begin{array}{l}\text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[\text{conjugate base]}}{[\text{weak acid]}}\\ \text{pH}{\text{= pK}}_{\text{a}}\text{ + log}\frac{[{\text{A}}^{\text{-}}\text{]}}{[\text{HA]}}\end{array}$

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Hydrolysis of salt of the weak base that is formed determines the pH of the solution at the equivalence point and it will fall in the pH region of 4 – 6.

After the equivalence point, the pH decreases to a level just greater than the pH of the strong acid titrant.

Equivalence point can be detected by the addition of indicator.

Answer to Problem 16.157QP

The conjugate pairs that can be used as a best indicator for,

1. (1) The titration of strong acid with a strong base is ${\text{B/BH}}^{+}$
2. (2) The titration of weak base with a strong acid is ${\text{HA/A}}^{-}$

Explanation of Solution

To Select: Among the conjugate pairs (i) ${\text{B/BH}}^{+}$ or (ii) ${\text{HA/A}}^{-}$ that can be used as a best indicator for the titration of strong acid with a strong base

The indicator for an acid-base titration is depends on the ${\text{pK}}_a$ value which overlaps with the pH of the equivalence point.

Since a strong acid-strong base titration gives a neutral solution at the point of equivalence, the choice of indicator must have ${\text{pK}}_a$ value close to 7.

Among the given conjugate pairs,  ${\text{B/BH}}^{+}$ and ${\text{HA/A}}^{-}$,

The ${\text{pK}}_a$ for is ${\text{BH}}^{+}$ 7.22

The ${\text{pK}}_a$ of $\text{HA}$ is 5.12

Therefore,  ${\text{B/BH}}^{+}$ indicator would be the appropriate indicator for the titration of strong acid with a strong base.

To Select: Among the conjugate pairs (i) ${\text{B/BH}}^{+}$ or (ii) ${\text{HA/A}}^{-}$ that can be used as a best indicator for the titration of weak base with a strong acid

The indicator for an acid-base titration is depends on the ${\text{pK}}_a$ value which overlaps with the pH of the equivalence point.

Since a weak base-strong acid titration gives a solution of the conjugate acid of the weak base at the point of equivalence, the choice of indicator must have ${\text{pK}}_a$ less than 7.

Among the given conjugate pairs,  ${\text{B/BH}}^{+}$ and ${\text{HA/A}}^{-}$,

The ${\text{pK}}_a$ for is ${\text{BH}}^{+}$ 7.22

The ${\text{pK}}_a$ of $\text{HA}$ is 5.12

Therefore, ${\text{HA/A}}^{-}$ indicator would be the appropriate indicator for the titration of weak base with a strong acid

Conclusion

The conjugate pairs that can be used as a best indicator for,

1. (1) The titration of strong acid with a strong base was found as ${\text{B/BH}}^{+}$
2. (2) The titration of weak base with a strong acid was found as ${\text{HA/A}}^{-}$