Chapter 11, Problem 11.9P

(a)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ and concentrations of ${\text{H}}_2\text{M}$, ${\text{HM}}^{-}$ and ${\text{M}}^{2-}$ in $0.100{\text{ M H}}_2\text{M}$ solution has to be calculated.

Concept Introduction:

$K_{\text{a2}}$ for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b2}}$ for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

Explanation of Solution

First equilibrium reaction for malonic acid that is designated as $K_{\text{a1}}$ is as follows:

  ${\text{H}}_2\text{M}\stackrel{K_{\text{a1}}}{\rightleftharpoons }{\text{HM}}^{-}+{\text{H}}^{+}$

The concentration of malonic acid is $0.100\text{ M}$. Consider concentration of ${\text{H}}^{+}$ to be $x$, thus value of $\left[{\text{HM}}^{-}\right]$ is also $x$ and that of $\left[{\text{H}}_2\text{M}\right]$ is $0.100-x$. The expression for equilibrium constant $K_{\text{a1}}$ for malonic acid is given as follows:

  $K_{\text{a1}}=\frac{\left[{\text{HM}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{H}}_2\text{M}\right]}$        (1)

Rearrange equation (1) for $\left[{\text{H}}^{+}\right]$.

  $\left[{\text{H}}^{+}\right]=\frac{K_{\text{a1}}\left[{\text{H}}_2\text{M}\right]}{\left[{\text{HM}}^{-}\right]}$        (2)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

$\left[{\text{H}}_2\text{M}\right]$ is concentration of ${\text{H}}_2\text{M}$.

Substitute $1.42\times {10}^{-3}$ for $K_{\text{a1}}$, $x$ for $\left[{\text{H}}^{+}\right]$, $x$ for $\left[{\text{HM}}^{-}\right]$ and $0.100-x$ for $\left[{\text{H}}_2\text{M}\right]$ in equation (2).

  $\begin{array}{c}x=\frac{\left(1.42\times {10}^{-3}\right)\left(0.100-x\right)}{x}\\ x^2=\left(1.42\times {10}^{-3}\right)\left(0.100-x\right)\\ x^2+\left(1.42\times {10}^{-3}\right)x-\left(1.42\times {10}^{-4}\right)=0\\ x=1.12\times {10}^{-2}\end{array}$

Therefore, value of $\left[{\text{H}}^{+}\right]$ is $1.12\times {10}^{-2}\text{ M}$, $\left[{\text{HM}}^{-}\right]$ is $1.12\times {10}^{-2}\text{ M}$ and $\left[{\text{H}}_2\text{M}\right]$ is $\text{0}\text{.089 M}$.

Second equilibrium reaction for malonic acid that is designated as $K_{\text{a2}}$ is as follows:

  ${\text{HM}}^{-}\stackrel{K_{\text{a2}}}{\rightleftharpoons }{\text{M}}^{2-}+{\text{H}}^{+}$

The expression for equilibrium constant $K_{\text{a2}}$ for malonic acid is given as follows:

  $K_{\text{a2}}=\frac{\left[{\text{M}}^{2-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HM}}^{-}\right]}$        (3)

Rearrange equation (3) for $\left[{\text{M}}^{2-}\right]$.

  $\left[{\text{M}}^{2-}\right]=\frac{K_{\text{a2}}\left[{\text{HM}}^{-}\right]}{\left[{\text{H}}^{+}\right]}$        (4)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

$\left[{\text{M}}^{2-}\right]$ is concentration of ${\text{M}}^{2-}$.

Substitute $2.01\times {10}^{-6}$ for $K_{\text{a2}}$, $1.12\times {10}^{-2}$ for $\left[{\text{H}}^{+}\right]$ and $1.12\times {10}^{-2}$ for $\left[{\text{HM}}^{-}\right]$ in equation (6).

  $\begin{array}{c}\left[{\text{M}}^{2-}\right]=\frac{\left(2.01\times {10}^{-6}\right)\left(1.12\times {10}^{-2}\right)}{\left(1.12\times {10}^{-2}\right)}\\ =2.01\times {10}^{-6}\text{ M}\end{array}$

Therefore, value of $\left[{\text{M}}^{2-}\right]$ is $2.01\times {10}^{-6}\text{ M}$.

(b)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ and concentrations of ${\text{H}}_2\text{M}$, ${\text{HM}}^{-}$ and ${\text{M}}^{2-}$ in $0.100\text{ M NaHM}$ solution has to be calculated.

Concept Introduction:

$K_{\text{a2}}$ for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b2}}$ for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

Explanation of Solution

The equilibrium reaction for malonic acid is as follows:

  $\text{NaHM}\rightleftharpoons {\text{Na}}^{+}+{\text{HM}}^{-}$

The expression used for calculation of concentration of ${\text{H}}^{+}$ is as follows:

  $\left[{\text{H}}^{+}\right]=\sqrt{\frac{K_{\text{a}1}{K}_{\text{a}2}\left[{\text{HM}}^{-}\right]+K_{\text{a1}}{K}_{\text{w}}}{K_{\text{a1}}+\left[{\text{HM}}^{-}\right]}}$        (5)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

$K_{\text{a1}}$ is first acid equilibrium constant.

$K_{\text{a}2}$ is second equilibrium constant.

$K_{\text{w}}$ is dissociation constant of water.

Substitute $1.42\times {10}^{-3}$ for $K_{\text{a1}}$, $2.01\times {10}^{-6}$ for $K_{\text{a2}}$, 0.100 for $\left[{\text{HM}}^{-}\right]$ and ${10}^{-14}$ for $K_{\text{w}}$ in equation (6).

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\sqrt{\frac{\left(1.42\times {10}^{-3}\right)\left(2.01\times {10}^{-6}\right)\left(0.100\right)+\left(1.42\times {10}^{-3}\right)\left({10}^{-14}\right)}{1.42\times {10}^{-3}+\left(0.100\right)}}\\ =\sqrt{\frac{2.85\times {10}^{-10}+1.42\times {10}^{-17}}{0.10142}}\\ =5.30\times {10}^{-5}\text{ M}\end{array}$

Therefore, value of $\left[{\text{H}}^{+}\right]$ is $5.30\times {10}^{-5}\text{ M}$.

The expression used to calculate $\text{pH}$ of $\left[{\text{H}}^{+}\right]$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (7)

Here,

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

Substitute $5.30\times {10}^{-5}$ for $\left[{\text{H}}^{+}\right]$ in equation (7).

  $\begin{array}{c}\text{pH}=-\log \left[5.30\times {10}^{-5}\right]\\ =-\log 5.30+5\log 10\\ =4.28\end{array}$

Hence, $\text{pH}$ of $\left[{\text{H}}^{+}\right]$ is 4.28.

First equilibrium reaction for malonic acid that is designated as $K_{\text{a1}}$ is as follows:

  ${\text{H}}_2\text{M}\stackrel{K_{\text{a1}}}{\rightleftharpoons }{\text{HM}}^{-}+{\text{H}}^{+}$

The expression for equilibrium constant $K_{\text{a1}}$ for malonic acid is given as follows:

  $K_{\text{a1}}=\frac{\left[{\text{HM}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{H}}_2\text{M}\right]}$        (8)

Rearrange equation (8) for $\left[{\text{H}}_2\text{M}\right]$.

  $\left[{\text{H}}_2\text{M}\right]=\frac{\left[{\text{HM}}^{-}\right]\left[{\text{H}}^{+}\right]}{K_{\text{a1}}}$        (9)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

$\left[{\text{H}}_2\text{M}\right]$ is concentration of ${\text{H}}_2\text{M}$.

Substitute $1.42\times {10}^{-3}$ for $K_{\text{a1}}$, $5.30\times {10}^{-5}$ for $\left[{\text{H}}^{+}\right]$ and 0.100 for $\left[{\text{HM}}^{-}\right]$  in equation (9).

  $\begin{array}{c}\left[{\text{H}}_2\text{M}\right]=\frac{\left(0.100\right)\left(5.30\times {10}^{-5}\right)}{1.42\times {10}^{-3}}\\ =3.73\times {10}^{-3}\text{ M}\end{array}$

Therefore, value of $\left[{\text{H}}_2\text{M}\right]$ is $3.73\times {10}^{-3}\text{ M}$.

Second equilibrium reaction for malonic acid that is designated as $K_{\text{a2}}$ is as follows:

  ${\text{HM}}^{-}\stackrel{K_{\text{a2}}}{\rightleftharpoons }{\text{M}}^{2-}+{\text{H}}^{+}$

The expression for equilibrium constant $K_{\text{a2}}$ for malonic acid is given as follows:

  $K_{\text{a2}}=\frac{\left[{\text{M}}^{2-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HM}}^{-}\right]}$        (10)

Rearrange equation (10) for $\left[{\text{M}}^{2-}\right]$.

  $\left[{\text{M}}^{2-}\right]=\frac{K_{\text{a2}}\left[{\text{HM}}^{-}\right]}{\left[{\text{H}}^{+}\right]}$        (11)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{H}}^{+}\right]$ is concentration of ${\text{H}}^{+}$.

$\left[{\text{M}}^{2-}\right]$ is concentration of ${\text{M}}^{2-}$.

Substitute $2.01\times {10}^{-6}$ for $K_{\text{a2}}$, $5.30\times {10}^{-5}$ for $\left[{\text{H}}^{+}\right]$ and 0.100 for $\left[{\text{HM}}^{-}\right]$ in equation (11).

  $\begin{array}{c}\left[{\text{M}}^{2-}\right]=\frac{\left(2.01\times {10}^{-6}\right)\left(0.100\right)}{\left(5.30\times {10}^{-5}\right)}\\ =3.9\times {10}^{-3}\text{ M}\end{array}$

Therefore, value of $\left[{\text{M}}^{2-}\right]$ is $3.9\times {10}^{-3}\text{ M}$.

(c)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ and concentrations of ${\text{H}}_2\text{M}$, ${\text{HM}}^{-}$ and ${\text{M}}^{2-}$ in $0.100{\text{ M Na}}_2\text{M}$ solution has to be calculated.

Concept Introduction:

$K_{\text{a2}}$ for reaction is defined as second dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b2}}$ for reaction is defined as second association constant for base. It helps to determine the amount of hydroxide ions in base.

Explanation of Solution

The equilibrium reaction for ${\text{Na}}_2\text{M}$ is as follows:

  ${\text{Na}}_2\text{M}\rightleftharpoons {\text{Na}}^{+}+{\text{M}}^{2-}$        (12)

Equilibrium reaction for ${\text{M}}^{2-}$ is as follows:

  $\begin{array}{l}{\text{M}}^{2-}+{\text{H}}_2\text{O}\stackrel{K_{\text{b1}}}{\rightleftharpoons }{\text{HM}}^{-}+{\text{OH}}^{-}\\ {\text{HM}}^{-}+{\text{H}}_2\text{O}\stackrel{K_{\text{b2}}}{\rightleftharpoons }{\text{H}}_2\text{M}+{\text{OH}}^{-}\end{array}$

The relation between $K_{a2}$ and $K_{\text{b1}}$ is as follows:

  $\left(K_{a2}\right)\left(K_{\text{b1}}\right)=K_{\text{w}}$        (13)

Rearrange equation (13) for $K_{\text{b1}}$.

  $K_{\text{b1}}=\frac{K_{\text{w}}}{K_{a2}}$        (14)

Here,

$K_{a2}$ is second acid equilibrium constant.

$K_{\text{b1}}$ is first base equilibrium constant.

$K_{\text{w}}$ is equilibrium constant for water.

Substitute $2.06\times {10}^{-6}$ for $K_{a2}$ and ${10}^{-14}$ for $K_{\text{w}}$ in equation (14).

  $\begin{array}{c}{K}_{\text{b1}}=\frac{{10}^{-14}}{2.06\times {10}^{-6}}\\ =4.85\times {10}^{-9}\end{array}$

Thus, value of $K_{\text{b1}}$ is $4.85\times {10}^{-9}$.

Consider concentration of ${\text{OH}}^{-}$ to be $x$, thus value of $\left[{\text{HM}}^{-}\right]$ is also $x$ and that of $\left[{\text{M}}^{2-}\right]$ is $0.100-x$. The expression for equilibrium constant $K_{\text{b1}}$ for malonic acid is given as follows:

  $K_{\text{b1}}=\frac{\left[{\text{HM}}^{-}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{M}}^{2-}\right]}$        (15)

Rearrange equation (15) for $\left[{\text{OH}}^{-}\right]$.

  $\left[{\text{OH}}^{-}\right]=\frac{K_{\text{b1}}\left[{\text{M}}^{2-}\right]}{\left[{\text{HM}}^{-}\right]}$        (16)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{OH}}^{-}\right]$ is concentration of ${\text{OH}}^{-}$.

$\left[{\text{M}}^{2-}\right]$ is concentration of ${\text{M}}^{2-}$.

$K_{\text{b1}}$ is first base equilibrium constant.

Substitute $4.85\times {10}^{-9}$ for $K_{\text{b1}}$, $x$ for $\left[{\text{OH}}^{-}\right]$, $x$ for $\left[{\text{HM}}^{-}\right]$ and $0.100-x$ for $\left[{\text{M}}^{2-}\right]$ in equation (16).

  $\begin{array}{c}x=\frac{\left(4.85\times {10}^{-9}\right)\left(0.100-x\right)}{x}\\ x^2=\left(4.85\times {10}^{-9}\right)\left(0.100-x\right)\\ x^2+\left(4.85\times {10}^{-9}\right)x-\left(4.85\times {10}^{-10}\right)=0\\ x=2.2\times {10}^{-5}\end{array}$

Therefore, value of $\left[{\text{OH}}^{-}\right]$ is $2.2\times {10}^{-5}\text{ M}$ and $\left[{\text{HM}}^{-}\right]$ is $2.2\times {10}^{-5}\text{ M}$.

The expression used for calculation of $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (17)

The expression used to calculate $K_{\text{w}}$ is as follows:

  $K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (18)

Rearrange equation (18) for $\left[{\text{H}}^{+}\right]$.

  $\left[{\text{H}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$        (19)

Substitute value of $\left[{\text{H}}^{+}\right]$ is equation (17).

  $\text{pH}=-\log \frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$        (20).

Substitute ${10}^{-14}$ for $K_{\text{w}}$ and $2.2\times {10}^{-5}$ for $\left[{\text{OH}}^{-}\right]$ in equation (20).

  $\begin{array}{c}\text{pH}=-\log \frac{{10}^{-14}}{2.2\times {10}^{-5}}\\ =9.34\end{array}$

The relation between $K_{a1}$ and $K_{\text{b2}}$ is as follows:

  $\left(K_{a1}\right)\left(K_{\text{b2}}\right)=K_{\text{w}}$        (21)

Rearrange equation (21) for $K_{\text{b2}}$.

  $K_{\text{b2}}=\frac{K_{\text{w}}}{K_{a1}}$        (22)

Here,

$K_{a1}$ is first acid equilibrium constant.

$K_{\text{b2}}$ is second equilibrium constant.

$K_{\text{w}}$ is equilibrium constant for water.

Substitute $1.42\times {10}^{-3}$ for $K_{a1}$ and ${10}^{-14}$ for $K_{\text{w}}$ in equation (22).

  $\begin{array}{c}{K}_{\text{b2}}=\frac{{10}^{-14}}{1.42\times {10}^{-3}}\\ =7.04\times {10}^{-12}\end{array}$

Thus, value of $K_{\text{b2}}$ is $7.04\times {10}^{-12}$.

The expression for equilibrium constant $K_{\text{b2}}$ is given as follows:

  $K_{\text{b2}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{H}}_2\text{M}\right]}{\left[{\text{HM}}^{-}\right]}$        (23)

Rearrange equation (23) for $\left[{\text{H}}_2\text{M}\right]$.

  $\left[{\text{H}}_2\text{M}\right]=\frac{K_{\text{b2}}\left[{\text{HM}}^{-}\right]}{\left[{\text{OH}}^{-}\right]}$        (24)

Here,

$\left[{\text{HM}}^{-}\right]$ is concentration of ${\text{HM}}^{-}$.

$\left[{\text{OH}}^{-}\right]$ is concentration of ${\text{OH}}^{-}$.

$\left[{\text{H}}_2\text{M}\right]$ is concentration of ${\text{H}}_2\text{M}$.

$K_{\text{b2}}$ is second base equilibrium constant.

Substitute $7.04\times {10}^{-12}$ for $K_{\text{b2}}$, $2.2\times {10}^{-5}$ for $\left[{\text{OH}}^{-}\right]$ and $2.2\times {10}^{-5}$ for $\left[{\text{HM}}^{-}\right]$ in equation (24).

  $\begin{array}{c}\left[{\text{H}}_2\text{M}\right]=\frac{\left(7.04\times {10}^{-12}\right)\left(2.2\times {10}^{-5}\right)}{\left(2.2\times {10}^{-5}\right)}\\ =7.04\times {10}^{-12}\end{array}$

Therefore, value of $\left[{\text{H}}_2\text{M}\right]$ is $7.04\times {10}^{-12}\text{ M}$.