Chapter 11, Problem 11.10P

(a)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ and concentrations of $\text{B}$, ${\text{BH}}^{+}$ and ${\text{BH}}_2^{2+}$ in $0.100\text{ M B}$ 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 compound $\text{B}$ is a dibasic compound thus, it requires two protons. Acceptance of one proton produces ${\text{BH}}^{+}$ and other produces ${\text{BH}}_2^{2+}$. The equilibrium reaction for addition of one proton is as follows:

  $\text{B}+{\text{H}}_2\text{O}\stackrel{K_{\text{b1}}}{\rightleftharpoons }{\text{BH}}^{+}+{\text{OH}}^{-}$

The ICE table for concentration of molecules in reaction is as follows:

  $\begin{array}{cccccccc}& \text{B}& +& {\text{H}}_{\text{2}}\text{O}& \stackrel{K_{\text{b}1}}{\rightleftharpoons }& {\text{BH}}^{\text{+}}& +& {\text{OH}}^{\text{-}}\\ \text{Initial,M}& 0.100& & & & 0& & 0\\ \text{Change,M}& -x& & & & x& & x\\ \text{Final,M}& 0.100-x& & & & x& & x\end{array}$

Concentration of $\text{B}$ is $0.100-x$ and of ${\text{BH}}^{+}$ and ${\text{OH}}^{-}$ is be $x$.

The expression for equilibrium constant $K_{\text{b1}}$ for dibasic acid is given as follows:

  $K_{\text{b1}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$        (1)

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

  $\left[{\text{BH}}^{+}\right]=\frac{K_{\text{b1}}\left[\text{B}\right]}{\left[{\text{OH}}^{-}\right]}$        (2)

Here,

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

$\left[\text{B}\right]$ is concentration of $\text{B}$.

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

$K_{\text{b1}}$ is basic equilibrium constant for dibasic compound.

Substitute $1.00\times {10}^{-5}$ for $K_{\text{b1}}$, $x$ for $\left[{\text{BH}}^{+}\right]$, $x$ for $\left[{\text{OH}}^{-}\right]$ and $0.100-x$ for $\left[\text{B}\right]$ in equation (2).

  $\begin{array}{c}x=\frac{\left(1.00\times {10}^{-5}\right)\left(0.100-x\right)}{x}\\ x^2=\left(1.00\times {10}^{-5}\right)\left(0.100-x\right)\\ x^2+\left(1.00\times {10}^{-5}\right)x-\left(1.00\times {10}^{-6}\right)=0\\ x=9.95\times {10}^{-4}\end{array}$

Therefore, value of $\left[{\text{BH}}^{+}\right]$ is $9.95\times {10}^{-4}$, $\left[{\text{OH}}^{-}\right]$ is $9.95\times {10}^{-4}$ and $\left[\text{B}\right]$ is $\text{0}\text{.09 M}$.

Second equilibrium reaction for dibasic compound $\text{B}$ that is designated as $K_{\text{b2}}$ is as follows:

  ${\text{BH}}^{+}\stackrel{K_{\text{b2}}}{\rightleftharpoons }{\text{BH}}_2^{2+}+{\text{OH}}^{-}$

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

  $K_{\text{b2}}=\frac{\left[{\text{BH}}_2^{2+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{BH}}^{+}\right]}$        (5)

Rearrange equation (5) for $\left[{\text{BH}}_2^{2+}\right]$.

  $\left[{\text{BH}}_2^{2+}\right]=\frac{K_{\text{b2}}\left[{\text{BH}}^{+}\right]}{\left[{\text{OH}}^{-}\right]}$        (6)

Here,

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

$\left[{\text{BH}}_2^{2+}\right]$ is concentration of ${\text{BH}}_2^{2+}$.

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

$K_{\text{b2}}$ is basic equilibrium constant for dibasic compound.

Substitute $1.00\times {10}^{-9}$ for $K_{\text{b2}}$, $9.95\times {10}^{-4}$ for $\left[{\text{BH}}^{+}\right]$, $9.95\times {10}^{-4}$ for $\left[{\text{OH}}^{-}\right]$ and in equation (6).

  $\begin{array}{c}\left[{\text{BH}}_2^{2+}\right]=\frac{\left(1.00\times {10}^{-9}\right)\left(9.95\times {10}^{-4}\right)}{\left(9.95\times {10}^{-4}\right)}\\ =1.00\times {10}^{-9}\end{array}$

Therefore, value of $\left[{\text{BH}}_2^{2+}\right]$ is $1.00\times {10}^{-9}\text{ M}$.

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

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

The expression for $K_{\text{w}}$ is as follows:

  $K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (8)

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

  $\left[{\text{H}}^{+}\right]=\frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$        (9)

Substitute value of $\left[{\text{H}}^{+}\right]$ is equation (7).

  $\text{pH}=-\log \frac{K_{\text{w}}}{\left[{\text{OH}}^{-}\right]}$        (10).

Substitute ${10}^{-14}$ for $K_{\text{w}}$ and $9.95\times {10}^{-4}$ for $\left[{\text{OH}}^{-}\right]$ in equation (10).

  $\begin{array}{c}\text{pH}=-\log \frac{{10}^{-14}}{9.95\times {10}^{-4}}\\ =11\end{array}$

Thus, $\text{pH}$ is 11, concentration of ${\text{BH}}^{+}$ is $9.95\times {10}^{-4}$, of $\text{B}$ is $\text{0}\text{.09 M}$ and of ${\text{BH}}_2^{2+}$ is $1.00\times {10}^{-9}\text{ M}$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{BH}}^{+}{\text{Br}}^{-}$ is an intermediate form from that removal of one proton produces $\text{B}$ and addition of one proton produces ${\text{BH}}_2^{2+}$. The equilibrium reaction is as follows:

  $\begin{array}{c}{\text{BH}}^{+}\rightleftharpoons \text{B}+{\text{H}}^{+}\\ {\text{BH}}^{+}+{\text{H}}^{+}\rightleftharpoons {\text{BH}}_2^{2+}\end{array}$

The formula to calculate value of $\text{pH}$ for given reaction is as follows:

  $\text{pH}=\frac{1}{2}\left(\text{p}{K}_{\text{b1}}+\text{p}{K}_{\text{b2}}\right)$        (11)

The expression for $\text{p}{K}_{\text{a}}$ is as follows:

  $\text{p}{K}_{\text{b}}=-\log K_{\text{b}}$        (12)

Substitute equation (12) in equation (11).

  $\text{pH}=\frac{1}{2}\left(-\text{log}{K}_{\text{b1}}+\left(-\log K_{\text{b2}}\right)\right)$        (13)

Substitute $1.00\times {10}^{-5}$ for $K_{\text{b1}}$ and $1.00\times {10}^{-9}\text{ M}$ for $K_{\text{b2}}$ in equation (13).

  $\begin{array}{c}\text{pH}=\frac{1}{2}\left(-\text{log}\left(1.00\times {10}^{-5}\right)+\left(-\log \left(1.00\times {10}^{-9}\text{ M}\right)\right)\right)\\ =\frac{1}{2}\left(5+9\right)\\ =7\end{array}$

The formula to calculate $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (14)

Rearrange equation (14) for $\left[{\text{H}}^{+}\right]$.

  $\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$        (15)

Substitute 7 for $\text{pH}$ in equation (15).

  $\left[{\text{H}}^{+}\right]={10}^{-\text{7}}\text{ M}$

The reaction for ${\text{BH}}_2^{2+}$ is given as follows:

  ${\text{BH}}_2^{2+}\rightleftharpoons {\text{BH}}^{+}+{\text{H}}^{+}$

The expression for equilibrium constant $K_{\text{b}1}$ is as follows:

  $K_{\text{b}1}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]}$        (16)

Rearrange equation (16) for $\left[{\text{BH}}_2^{2+}\right]$.

  $\left[{\text{BH}}_2^{2+}\right]=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{H}}^{+}\right]}{K_{\text{b}1}}$        (17)

Substitute $1.00\times {10}^{-5}$ for $K_{\text{b1}}$, ${10}^{-\text{7}}$ for $\left[{\text{H}}^{+}\right]$ and 0.100 for $\left[{\text{BH}}^{+}\right]$ in equation (17).

  $\begin{array}{c}\left[{\text{BH}}_2^{2+}\right]=\frac{\left(0.100\right)\left({10}^{-\text{7}}\right)}{1.00\times {10}^{-5}}\\ =1.0\times {10}^{-3}\text{ M}\end{array}$

The reaction for ${\text{BH}}^{+}$ is given as follows:

  ${\text{BH}}^{+}\rightleftharpoons \text{B}+{\text{H}}^{+}$

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

  $K_{\text{b2}}=\frac{\left[\text{B}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{BH}}^{+}\right]}$        (18)

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

  $\left[\text{B}\right]=\frac{K_{\text{b2}}\left[{\text{BH}}^{+}\right]}{\left[{\text{H}}^{+}\right]}$        (19)

Substitute $1.00\times {10}^{-9}$ for $K_{\text{b2}}$, ${10}^{-\text{7}}$ for $\left[{\text{H}}^{+}\right]$ and 0.100 for $\left[{\text{BH}}^{+}\right]$ in equation (19).

  $\begin{array}{c}\left[\text{B}\right]=\frac{\left(1.00\times {10}^{-9}\right)\left(0.100\right)}{{10}^{-\text{7}}}\\ =1.0\times {10}^{-3}\end{array}$

Thus, $\text{pH}$ is 7, concentration of ${\text{BH}}_2^{2+}$ is $1.0\times {10}^{-3}\text{ M}$, of ${\text{BH}}^{+}$ is $0.100\text{ M}$ and $\text{B}$ is $1.0\times {10}^{-3}\text{ M}$.

(c)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ and concentrations of $\text{B}$, ${\text{BH}}^{+}$ and ${\text{BH}}_2^{2+}$ in $0.100{\text{ M BH}}_2^{2+}{\left({\text{Br}}^{-}\right)}_2$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The compound $\text{B}$ is a dibasic compound thus, it requires two protons. Acceptance of one proton produces ${\text{BH}}^{+}$ and other produces ${\text{BH}}_2^{2+}$. The equilibrium reaction for addition of one proton is as follows:

  ${\text{BH}}_2^{2+}\stackrel{K_1}{\rightleftharpoons }{\text{BH}}^{+}+{\text{H}}^{+}$

The ICE table for concentration of molecules in reaction is as follows:

$\begin{array}{cccccc}& {\text{BH}}_2^{2+}& \stackrel{K_1}{\rightleftharpoons }& {\text{BH}}^{+}& +& {\text{H}}^{+}\\ \text{Initial}& 0.100& & 0& & 0\\ \text{Change}& -x& & x& & x\\ \text{Final}& 0.100-x& & x& & x\end{array}$

Concentration of ${\text{BH}}_2^{2+}$ is $0.100-x$ and of ${\text{BH}}^{+}$ and ${\text{H}}^{+}$ is be $x$.

The expression for equilibrium constant $K_{\text{b1}}$ for dibasic acid is given as follows:

  $K_{\text{b1}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]}$        (20)

Here,

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

$\left[{\text{BH}}_2^{2+}\right]$ is concentration of ${\text{BH}}_2^{2+}$.

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

$K_{\text{b1}}$ is basic equilibrium constant for dibasic compound.

Substitute $1.00\times {10}^{-5}$ for $K_{\text{b1}}$, $x$ for $\left[{\text{BH}}^{+}\right]$, $x$ for $\left[{\text{H}}^{+}\right]$ and $0.100-x$ for $\left[{\text{BH}}_2^{2+}\right]$ in equation (20).

  $\begin{array}{c}1.00\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.100-x}\\ x^2+\left(1.00\times {10}^{-5}\right)x-\left(1.00\times {10}^{-6}\right)=0\end{array}$

Sove above equation for $x$.

  $x=9.95\times {10}^{-4}$

Therefore, value of $\left[{\text{BH}}^{+}\right]$ is $9.95\times {10}^{-4}$, $\left[{\text{H}}^{+}\right]$ is $9.95\times {10}^{-4}$ and $\left[{\text{BH}}_2^{2+}\right]$ is $\text{0}\text{.09 M}$.

Second equilibrium reaction for hydrolysis of dibasic compound $\text{B}$ that is designated as $K_{\text{b2}}$ is as follows:

  ${\text{BH}}^{+}\stackrel{K_{\text{b2}}}{\rightleftharpoons }\text{B}+{\text{H}}^{+}$

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

  $K_{\text{b2}}=\frac{\left[\text{B}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{BH}}^{+}\right]}$        (21)

Rearrange equation (21) for $\left[\text{B}\right]$.

  $\left[\text{B}\right]=\frac{K_{\text{b2}}\left[{\text{BH}}^{+}\right]}{\left[{\text{H}}^{+}\right]}$        (22)

Substitute $1.00\times {10}^{-9}$ for $K_{\text{b2}}$, $9.95\times {10}^{-4}$ for $\left[{\text{BH}}^{+}\right]$, $9.95\times {10}^{-4}$ for $\left[{\text{H}}^{+}\right]$ in equation (22).

  $\begin{array}{c}\left[\text{B}\right]=\frac{\left(1.00\times {10}^{-9}\right)\left(9.95\times {10}^{-4}\right)}{\left(9.95\times {10}^{-4}\right)}\\ =1.00\times {10}^{-9}\end{array}$

Therefore, value of $\left[\text{B}\right]$ is $1.00\times {10}^{-9}\text{ M}$.

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

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$        (23)

Substitute $9.95\times {10}^{-4}$ for $\left[{\text{H}}^{+}\right]$ in equation (23).

  $\begin{array}{c}\text{pH}=-\log \left(9.95\times {10}^{-4}\right)\\ =-\log 9.95+4\\ =-0.998+4\\ =3\end{array}$

Thus, concentration of ${\text{BH}}^{+}$ is $9.95\times {10}^{-4}$, of ${\text{BH}}_2^{2+}$ is $\text{0}\text{.09 M}$ and of $\text{B}$ is $1.00\times {10}^{-9}\text{ M}$ and $\text{pH}$ is 3.