Chapter 16, Problem 16.70QP

(a)

Interpretation Introduction

Interpretation:

The pH for the given solutions has to be calculated

Concept Information:

Strong base and weak base:

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water.  Weak bases partially dissociates into its constituent ions.

According to Bronsted-Lowry, strong base is a good proton acceptor whereas weak base is a poor proton acceptor

Since, the ionization of a weak base is incomplete; it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base $\text{B}$ is given by the below equation.

${\text{B}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HB}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where,

$K_b$ is base ionization constant,

$[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion

$[{\text{HB}}^{\text{+}}\text{]}$  is concentration of conjugate acid

$[\text{B]}$ is concentration of the base

$\text{pOH}$definition:

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The pH of the given solutions

To calculate the pH of 0.10 M ${\text{NH}}_{\text{3}}$

Answer to Problem 16.70QP

Answer

The pH of the given solution (a) is 11.11

Explanation of Solution

Record the given data

A 0.10-M solution of ammonia at $\text{25}{}^{\text{o}}\text{C}$.

From the concentration and ${\text{K}}_b$ of the given ammonia, the hydroxide ion concentration can be found.  From hydroxide ion concentration, the pOH and finally pH can be calculated

Calculation of ${\text{K}}_b$

The equilibrium table for ammonia can be constructed as follows,

	${\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	$0.10$ $-x$ $0.10-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{NH}}_{\text{4}}{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{NH}}_{\text{3}}]}\\ \text{The base ionization constant for ammonia is 1}\text{.8}\times {\text{10}}^{\text{-5}}\\ \text{1}\text{.8}\times {\text{10}}^{\text{-5}}=\frac{x^2}{(0.10-x)}\\ \text{Assuming (0}\text{.10-}x\text{)}\approx 0.10,\text{ we have:}\\ \text{1}\text{.8}\times {\text{10}}^{\text{-5}}=\frac{x^2}{(0.10)}\\ x=1.3\times {10}^{-3}M\end{array}$

Therefore, the concentration of hydroxide ion is $1.3\times {10}^{-3}M$

Calculation of pOH:

The pOH can be calculated as follows,

$\begin{array}{l}{\text{pOH=-log[OH}}^{\text{-}}]\\ =-\log (1.3\times {10}^{-3})\\ =2.89\end{array}$

Calculation of pH

The pH can be calculated using the following formula as follows,

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pH=14}\text{.00-2}\text{.89}\\ \text{=11}\text{.11}\end{array}$

Therefore, the pH of 0.10 M ammonia solution is 11.11

(b)

Interpretation Introduction

Interpretation:

The pH for the given solutions has to be calculated

Concept Information:

Strong base and weak base:

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water.  Weak bases partially dissociates into its constituent ions.

According to Bronsted-Lowry, strong base is a good proton acceptor whereas weak base is a poor proton acceptor

Since, the ionization of a weak base is incomplete; it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base $\text{B}$ is given by the below equation.

${\text{B}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HB}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where,

$K_b$ is base ionization constant,

$[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion

$[{\text{HB}}^{\text{+}}\text{]}$  is concentration of conjugate acid

$[\text{B]}$ is concentration of the base

$\text{pOH}$definition:

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between $\text{pH}$ and $\text{pOH}$

$\text{pOH}$ is similar to $\text{pH}$.  The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The pH of the given solutions

To calculate the pH of 0.050 M pyridine

Answer to Problem 16.70QP

Answer

The pH of the given solution (b) is 8.96

Explanation of Solution

Record the given datas

A 0.050 M pyridine at $\text{25}{}^{\text{o}}\text{C}$.

From the concentration and ${\text{K}}_b$ of the given pyridine, the hydroxide ion concentration can be found.  From hydroxide ion concentration, the pOH and finally pH can be calculated

Calculation of ${\text{K}}_b$

The equilibrium table for ammonia can be constructed as follows,

	${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{N}}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	$0.050$ $-x$ $0.050-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}]}\\ \text{The base ionization constant for ammonia is 1}\text{.7}\times {\text{10}}^{\text{-9}}\\ \text{1}\text{.7}\times {\text{10}}^{\text{-9}}=\frac{x^2}{(0.050-x)}\\ \text{Assuming (0}\text{.050-}x\text{)}\approx 0.050,\text{ we have:}\\ \text{1}\text{.8}\times {\text{10}}^{\text{-5}}=\frac{x^2}{(0.050)}\\ x=3\times {10}^{-4}M\end{array}$

Therefore, the concentration of hydroxide ion is $3\times {10}^{-4}M$

Calculation of pOH:

The pOH can be calculated as follows,

$\begin{array}{l}{\text{pOH=-log[OH}}^{\text{-}}]\\ =-\log (3\times {10}^{-4})\\ =3.52\end{array}$

Calculation of pH

The pH can be calculated using the following formula as follows,

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pH=14}\text{.00-3}\text{.52}\\ \text{=10}\text{.48}\end{array}$

Therefore, the pH of 0.050 M pyridine solution is 10.48