Chapter 14, Problem 107CWP

(a)

Interpretation Introduction

Interpretation: The four titrations in order of increasing $\text{pH}$ at the halfway point to equivalence (lowest to highest) and at the equivalence point are to be ranked. The titration requires the largest volume of titrant $\text{HCl}$ or $\text{NaOH}$ is to be stated.

Concept introduction: Titration or acid/base titration is a chemical process of neutralization in which a titrant that is the solution of known concentration reacts with a solution of unknown concentration.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

At the half equivalence point the concentration of titrant is just half of the initial amount.

For the given value of dissociation constant the $\text{pH}=\text{p}{K}_{\text{a}}$.

To determine: The rank of four titrations in order of increasing $\text{pH}$ at the halfway point to equivalence (lowest to highest).

Answer to Problem 107CWP

Answer

The rank of four titrations in order of increasing $\text{pH}$ at the halfway point to equivalence (lowest to highest) is $\underset{\_}{II<IV<III<I}$.

Explanation of Solution

Explanation

Given

The four titrations are given as,

1. I. $150\text{ml}$ of $0.2\text{M}{\text{NH}}_3\left(K_{\text{b}}=1.8\times {10}^{-5}\right)$ by $0.2\text{M}\text{HCl}$.
2. II. $150\text{ml}$ of $0.2\text{M}\text{HCl}$ by $0.2\text{M}\text{NaOH}$.
3. III. $150\text{ml}$ of $0.2\text{M}\text{HOCl}\left(K_{\text{a}}=3.5\times {10}^{-8}\right)$ by $0.2\text{M}\text{NaOH}$.
4. IV. $150\text{ml}$ of $0.2\text{M}\text{HF}\left(K_{\text{a}}=7.2\times {10}^{-4}\right)$ by $0.2\text{M}\text{NaOH}$.

For the first acid-base titration the $\text{pH}$ is calculated by the following,

To get the $\text{pH}$ the Henderson equation is written as,

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{Conjugate}\text{base}\right]}{\left[\text{Weak}\text{acid}\right]}$

In the titration of conjugate base with weak acid at the half equivalence point the concentration of both becomes same therefore, the Henderson equation written as,

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{Conjugate}\text{base}\right]}{\left[\text{Weak}\text{acid}\right]}\\ \text{pH}=\text{p}{K}_{\text{a}}+\log \left(1\right)\\ \text{pH}=\text{p}{K}_{\text{a}}\end{array}$ (1)

The $\text{p}{K}_{\text{b}}$ of the base is calculated by the formula,

$\text{p}{K}_{\text{b}}=-\log \left(K_{\text{b}}\right)$

Where,

$K_{\text{b}}$ is the base dissociation constant.

Substitute the value of $K_{\text{b}}$ in the above equation.

$\begin{array}{l}\text{p}{K}_{\text{b}}=-\log \left(K_{\text{b}}\right)\\ \text{p}{K}_{\text{b}}=-\log \left(1.8\times {10}^{-5}\right)\\ \text{p}{K}_{\text{b}}=-\log \left(1.8\right)+5\\ \text{p}{K}_{\text{b}}=4.74\end{array}$

The value of ${\text{pK}}_{\text{a}}$ is calculated by the formula,

$\text{p}{K}_{\text{a}}+\text{p}{K}_{\text{b}}=14$

Substitute the calculated value of $\text{p}{K}_{\text{b}}$ in the above equation.

$\begin{array}{c}\text{p}{K}_{\text{a}}+\text{p}{K}_{\text{b}}=14\\ \text{p}{K}_{\text{a}}=14-\text{p}{K}_{\text{b}}\\ \text{p}{K}_{\text{a}}=14-4.74\\ \text{p}{K}_{\text{a}}=9.26\end{array}$

Substitute the calculated value of $\text{p}{K}_a$ in the equation (1).

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}\\ \text{pH}=\text{9}\text{.26}\end{array}$

For the second acid-base titration the $\text{pH}$ is calculated as,

For the strong acid-base titration, the $\text{pH}$ at the half equivalence point is the concentration of titrant which is just half of the initial amount of the concentration.

Therefore, for the second titration the half way equivalence point is calculated as,

$\begin{array}{l}\left[\text{HI}\right]=\frac{\text{Initial}\text{concentration}}{2}\\ \left[\text{HI}\right]=\frac{0.2\text{M}}{2}\\ \left[\text{HI}\right]=0.1\text{M}\end{array}$

The given titrant is an acid and the concentration of an acid is equal to the hydrogen ion concentration. Therefore,

$\left[{\text{H}}^{\text{+}}\right]=0.1$

The $\text{pH}$ of the base is calculated by the formula,

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

Where,

$\left[{\text{H}}^{\text{+}}\right]$ is the Hydrogen ion concentration.

Substitute the value of Hydrogen ion concentration in the above equation.

$\begin{array}{l}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ \text{pH}=-\log \left(0.1\right)\\ \text{pH}=-\log \left({10}^{-1}\right)\\ \text{pH}=1\end{array}$

For the third conjugate acid-base titration the $\text{pH}$ is calculated as,

The $\text{pH}$ at half equivalence point is calculated by the equation (1).

$\text{pH}=\text{p}{K}_{\text{a}}$ (2)

The value $\text{p}{K}_{\text{a}}$ is calculated as,

$\begin{array}{l}\text{p}{K}_{\text{a}}=-\log \left(K_{\text{a}}\right)\\ \text{p}{K}_{\text{a}}=-\log \left(K_{\text{a}}\right)\\ \text{p}{K}_{\text{a}}=-\log \left(3.5\times {10}^{-8}\right)\\ \text{p}{K}_{\text{a}}=7.45\end{array}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in the equation (2).

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}\\ \text{pH}=\text{7}\text{.45}\end{array}$

For the fourth conjugate acid-base titration the $\text{pH}$ is calculated as,

For the given value of acid dissociation constant,

$\text{pH}=\text{p}{K}_{\text{a}}$

The value $\text{p}{K}_{\text{a}}$ is calculated as,

$\begin{array}{l}\text{p}{K}_{\text{a}}=-\log \left(K_{\text{a}}\right)\\ \text{p}{K}_{\text{a}}=-\log \left(K_{\text{a}}\right)\\ \text{p}{K}_{\text{a}}=-\log \left(7.2\times {10}^{-4}\right)\\ \text{p}{K}_{\text{a}}=3.14\end{array}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in the equation (2).

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}\\ \text{pH}=\text{3}\text{.14}\end{array}$

The calculated $\text{pH}$ values at the halfway point of equivalence for the given four titrations are given as,

1. i. $9.26$
2. ii. $1$
3. iii. $7.45$
4. iv. $3.14$

Therefore, the increasing order of $\text{pH}$ is given as,

$\underset{\_}{II<IV<III<I}$.

(b)

Interpretation Introduction

Interpretation: The four titrations in order of increasing $\text{pH}$ at the halfway point to equivalence (lowest to highest) and at the equivalence point are to be ranked. The titration requires the largest volume of titrant $\text{HCl}$ or $\text{NaOH}$ is to be stated.

Concept introduction: Titration or acid/base titration is a chemical process of neutralization in which a titrant that is the solution of known concentration reacts with a solution of unknown concentration.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

At the half equivalence point the concentration of titrant is just half of the initial amount.

For the given value of dissociation constant the $\text{pH}=\text{p}{K}_{\text{a}}$.

To determine: The rank of four titrations in order of increasing $\text{pH}$ at the equivalence point.

Answer to Problem 107CWP

Answer

The rank of four titrations in order of increasing $\text{pH}$ at the equivalence is.

$\underset{\_}{I<II<IV<III}$.

Explanation of Solution

Explanation

For the first acid-base titration the $\text{pH}$ is calculated by the following,

In the acid-base titration the aqueous solution of ammonia in the reaction form is given as,

${\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{3}}\text{O}\left(aq\right)\to {\text{NH}}_{\text{4}}^{\text{+}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The volume of $\text{HCl}$ is calculated by the formula,

$M_1{V}_1=M_2{V}_2$

Where,

• $M_1$ is the concentration of ammonia.
• $V_1$ is the volume of ammonia.
• $M_2$ is the concentration of the acid.
• $V_2$ is the volume of the acid.

Substitute the values in the above formula.

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.2\text{M}\times 150\text{ml}=0.2\text{M}\times V_2\\ V_2=150\text{ml}\end{array}$

At the equivalence point the volume of solution containing ${\text{NH}}_{\text{4}}^{\text{+}}$ becomes $150+150=300\text{ml}$.

Moles of acid and base is calculated by the formula,

$n=C\times V$

Where,

• $C$ is the concentration.
• $V$ is the volume.

Substitute the values in the above formula.

For Ammonia,

$\begin{array}{l}n=C\times V\\ n=0.1\times \frac{150}{1000}\text{L}\\ n=\text{0}\text{.015}\text{mol}\end{array}$

For acid,

$\begin{array}{l}n=C\times V\\ n=0.1\text{M}\times \frac{150}{1000}\text{L}\\ n=0.015\text{mol}\end{array}$

Molarity $\left(M\right)$ of ${\text{NH}}_{\text{4}}^{\text{+}}$ is calculated by the formula,

$M=\frac{n}{V}$

Substitute the values.

$\begin{array}{l}M=\frac{n}{V}\\ M=\frac{0.015}{0.300}\\ M=0.05\text{M}\end{array}$

The relation between $K_{\text{a}}$ and $K_{\text{b}}$ is given as,

$K_{\text{a}}\times K_{\text{b}}=1\times {10}^{-14}$

Substitute the value of $K_{\text{b}}$.

$\begin{array}{c}{K}_{\text{a}}\times K_{\text{b}}=1\times {10}^{-14}\\ K_{\text{a}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}\\ K_{\text{a}}=5.55\times {10}^{-10}\end{array}$

The concentration of products is suppose to be $x$. The change in concentration for the titration reaction is given as,

$\begin{array}{cccccccc}& {\text{NH}}_4^{+}& +& {\text{H}}_2\text{O}& \to & {\text{NH}}_3& +& {\text{H}}_3{\text{O}}^{+}\\ \text{Initial}\text{concentration}& 0.2& & -& & 0& & 0\\ \text{Equilibrium}\text{concentration}& 0.05& & -& & x& & x\end{array}$

The concentration of For the given acid-base titration the acid dissociation constant is written as,

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_4^{+}\right]}$

Substitute the values in the above formula.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{NH}}_{\text{3}}\right]}{\left[{\text{NH}}_4^{+}\right]}\\ 5.55\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{0.05}\\ x^2=0.27\times {10}^{-10}\\ x=5.26\times {10}^{-6}\end{array}$

The calculated value of $x$ is the concentration of Hydrogen ion. Therefore the $\text{pH}$ is calculated as,

$\begin{array}{l}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ \text{pH}=-\log \left(5.26\times {10}^{-6}\right)\\ \text{pH}=-\log \left(5.26\right)+6\\ \text{pH}=5.27\end{array}$

For the second acid-base titration the $\text{pH}$ is calculated by the following,

In the strong acid-base titration the reaction of neutralization is given as, $\text{KOH}+\text{HCl}\to \text{KCl}+{\text{H}}_{\text{2}}\text{O}$

The given acid is a strong acid and base is also a strong base. Therefore, the complete neutralization process takes place. Hence, the $\text{pH}$ is $7$

For the third acid-base titration the $\text{pH}$ is calculated by the following,

In the acid-base titration the reaction is given as,

$\text{HOCl}+\text{NaOH}\to \text{NaOCl}+{\text{H}}_{\text{2}}\text{O}$

The volume of $\text{NaOH}$ is calculated by the formula,

$M_1{V}_1=M_2{V}_2$

Substitute the values in the above formula.

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.2\text{M}\times 150\text{ml}=0.2\text{M}\times V_2\\ V_2=150\text{ml}\end{array}$

At the equivalence point the volume of solution containing $\text{NOCl}$ becomes $150+150=300\text{ml}$.

Moles of acid and base is calculated by the formula,

$n=C\times V$

Where,

• $C$ is the concentration.
• $V$ is the volume.

Substitute the values in the above formula.

For $\text{HOCl}$,

$\begin{array}{l}n=C\times V\\ n=0.1\times \frac{150}{1000}\text{L}\\ n=\text{0}\text{.015}\text{mol}\end{array}$

For $\text{NaOH}$ ,

$\begin{array}{l}n=C\times V\\ n=0.1\text{M}\times \frac{150}{1000}\text{L}\\ n=0.015\text{mol}\end{array}$

Molarity $\left(M\right)$ of $\text{NaOCl}$ is calculated by the formula,

$\begin{array}{l}M=\frac{n}{V}\\ M=\frac{0.015}{0.300}\\ M=0.05\text{M}\end{array}$

The value of $K_{\text{b}}$.is calculated as,

$\begin{array}{c}{K}_{\text{a}}\times K_{\text{b}}=1\times {10}^{-14}\\ K_{\text{b}}=\frac{1\times {10}^{-14}}{3.5\times {10}^{-8}}\\ K_{\text{b}}=2.86\times {10}^{-7}\end{array}$

The concentration of products is suppose to be $y$. The change in concentration for the reaction is given as,

$\begin{array}{cccccccc}& \text{NaOCl}\left(aq\right)& +& {\text{H}}_2\text{O}\left(l\right)& \to & \text{HOCl}\left(aq\right)& +& {\text{OH}}^{-}\\ \text{Initial}\text{concentration}& 0.05& & -& & 0& & 0\\ \text{Equilibrium}\text{concentration}& 0.05-y& & -& & y& & y\end{array}$

The concentration of For the given acid-base titration the acid dissociation constant is written as,

$K_{\text{a}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{NaOCl}\right]}$

Substitute the values in the above formula.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{NaOCl}\right]}\\ 2.86\times {10}^{-7}=\frac{\left(y\right)\left(y\right)}{\left(0.05-y\right)}\\ y^2=2.86\times {10}^{-7}\times \left(0.05-y\right)\\ y^2+2.86\times {10}^{-7}y-1.43\times {10}^{-8}=0\end{array}$

Simplify the above quadratic equation.

$y=1.19\times {10}^{-4}$

The calculated value of $y$ is the concentration of Hydroxide ion that is the $\left[{\text{OH}}^{\text{-}}\right]$.

Therefore the $\text{pOH}$ is calculated as,

$\begin{array}{l}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ \text{pOH}=-\log \left(1.19\times {10}^{-4}\right)\\ \text{pOH}=-\log \left(1.19\right)+4\\ \text{pOH}=3.92\end{array}$

The $\text{pH}$ is calculated as,

$\begin{array}{l}\text{pH}=14-\text{pOH}\\ \text{pH}=14-\text{3}\text{.92}\\ \text{pH}=10.08\end{array}$

For the fourth acid-base titration the $\text{pH}$ is calculated by the following,

In the acid-base titration the reaction is given as,

$\text{HF}+\text{NaOH}\to \text{NaF}+{\text{H}}_{\text{2}}\text{O}$

The volume of $\text{NaOH}$ is calculated by the formula,

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ 0.2\text{M}\times 150\text{ml}=0.2\text{M}\times V_2\\ V_2=150\text{ml}\end{array}$

At the equivalence point the volume of solution containing $\text{NOCl}$ becomes $150+150=300\text{ml}$.

Moles of acid and base is calculated by the formula,

For $\text{HF}$,

$\begin{array}{l}n=C\times V\\ n=0.2\times \frac{150}{1000}\text{L}\\ n=\text{0}\text{.030}\text{mol}\end{array}$

For $\text{NaOH}$ ,

$\begin{array}{l}n=C\times V\\ n=0.2\text{M}\times \frac{150}{1000}\text{L}\\ n=0.030\text{mol}\end{array}$

Molarity $\left(M\right)$ of $\text{NaF}$ is calculated by the formula,

$\begin{array}{l}M=\frac{n}{V}\\ M=\frac{0.015}{0.300}\\ M=0.05\text{M}\end{array}$

The value of $K_{\text{b}}$.is calculated as,

$\begin{array}{c}{K}_{\text{a}}\times K_{\text{b}}=1\times {10}^{-14}\\ K_{\text{b}}=\frac{1\times {10}^{-14}}{7.2\times {10}^{-4}}\\ K_{\text{b}}=1.39\times {10}^{-11}\end{array}$

The concentration of products is suppose to be $y$. The change in concentration for the reaction is given as,

$\begin{array}{cccccccc}& \text{NaF}\left(aq\right)& +& {\text{H}}_2\text{O}\left(l\right)& \to & \text{HF}\left(aq\right)& +& {\text{OH}}^{-}\\ \text{Initial}\text{concentration}& 0.2& & -& & 0& & 0\\ \text{Equilibrium}\text{concentration}& 0.05& & -& & y& & y\end{array}$

The concentration of For the given acid-base titration the base dissociation constant is written as,

$K_{\text{b}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{NaOCl}\right]}$

Substitute the values in the above formula.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[\text{HOCl}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{NaOCl}\right]}\\ 1.39\times {10}^{-11}=\frac{\left(y\right)\left(y\right)}{\left(0.05\right)}\\ y^2=0.695\times {10}^{-12}\\ y=8.3\times {10}^{-6}\end{array}$

The calculated value of $y$ is the concentration of Hydroxide ion that is the $\left[{\text{OH}}^{\text{-}}\right]$.

Therefore the $\text{pOH}$ is calculated as,

$\begin{array}{l}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ \text{pOH}=-\log \left(8.3\times {10}^{-7}\right)\\ \text{pOH}=-\log \left(8.3\right)+7\\ \text{pOH}=6.08\end{array}$

The $\text{pH}$ is calculated as,

$\begin{array}{l}\text{pH}=14-\text{pOH}\\ \text{pH}=14-6.08\\ \text{pH}=7.91\end{array}$

The calculated $\text{pH}$ values of the equivalence for the given four titrations are given as,

1. i. $5.27$
2. ii. $7$
3. iii. $10.08$
4. iv. $7.91$

Therefore, the increasing order of $\text{pH}$ is given as,

$\underset{\_}{I<II<IV<III}$.

(c)

Interpretation Introduction

Interpretation: The four titrations in order of increasing $\text{pH}$ at the halfway point to equivalence (lowest to highest) and at the equivalence point are to be ranked. The titration requires the largest volume of titrant $\text{HCl}$ or $\text{NaOH}$ is to be stated.

Concept introduction: Titration or acid/base titration is a chemical process of neutralization in which a titrant that is the solution of known concentration reacts with a solution of unknown concentration.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

At the half equivalence point the concentration of titrant is just half of the initial amount.

For the given value of dissociation constant the $\text{pH}=\text{p}{K}_{\text{a}}$.

To determine: The titration that requires the largest volume of titrant to reach the equivalence point.

Answer to Problem 107CWP

All the titration requires the same volume.

Explanation of Solution

Explanation

All the given titrants are of same concentration and the given value of volume is also same. Therefore, the volume required of acid and base in each titration is same.

Conclusion

Conclusion

All the given questions based on $\text{pH}$ have been rightfully stated