Chapter 16, Problem 16.70QP

(a)

Interpretation Introduction

To determine: The $\text{pH}$ value of a $0.0133\text{M}$ solution of piperazine.

Interpretation: The $\text{pH}$ value of a $0.0133\text{M}$ solution of piperazine is to be calculated.

Concept introduction: The $\text{pH}$ value of a compound is defined as the log of the concentration of hydronium ion released by the compound upon dissociation. The value of $\text{pH}$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $\text{pH}$ is $7$.

Answer to Problem 16.70QP

The $\text{pH}$ of a $0.0133\text{M}$ solution of piperazine is $\underset{\_}{10.92}$.

Explanation of Solution

The molar concentration of piperazine is $0.0133\text{M}$.

The values of $K_{\text{b1}}$ and $K_{\text{b2}}$ for piperazine is $5.38\times {10}^{-5}$ and $2.15\times {10}^{-9}$.

The first ionization reaction of piperazine is,

$\text{B}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{BH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Here, $\text{B}$ is piperazine.

The ICE Table for ionization of piperazine is,

$\begin{array}{cccc}& \left[\text{B}\right]\text{M}& \left[{\text{BH}}^{+}\right]\text{M}& \left[{\text{OH}}^{-}\right]\text{M}\\ \text{Intial}& 0.0133& 0& 0\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 0.0133-x& x& x\end{array}$

The equilibrium constant for the above reaction is,

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

Where,

• $K_{\text{b2}}$ is equilibrium constant for the above reaction.
• $\left[{\text{BH}}_2{}^{+}\right]$ is concentration of piperazine ion.
• $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxide ion.
• $\left[\text{B}\right]$ is concentration of piperazine.
• $x$ is assumed concentration of hydroxide ion.

Substitute the value of equilibrium concentration from to table to above expression.

$\begin{array}{c}{K}_{\text{b1}}=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}\\ 5.38\times {10}^{-5}=\frac{{\left(x\right)}^2}{\left(0.0133-x\right)}\\ x=8.46\times {10}^{-4}\end{array}$

Hence, the concentration of hydroxide ion is $8.46\times {10}^{-4}$.

The concentrations of piperazine and piperazine ion by ICE table are $1.24\times {10}^{-2}$ and $8.46\times {10}^{-4}$.

The second ionization reaction of piperazine is,

${\text{BH}}^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{BH}}_2{}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Here, $\text{B}$ is piperazine.

The ICE Table for ionization of piperazine is,

$\begin{array}{cccc}& \left[{\text{BH}}^{+}\right]\text{M}& \left[{\text{BH}}_2{}^{+}\right]\text{M}& \left[{\text{OH}}^{-}\right]\text{M}\\ \text{Intial}& 8.46\times {10}^{-4}& 0& 8.46\times {10}^{-4}\\ \text{Change}& -x& +x& +x\\ \text{Equlibrium}& 8.46\times {10}^{-4}-x& x& 8.46\times {10}^{-4}+x\end{array}$

The equilibrium constant for the above reaction is,

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

Where,

• $K_{\text{b1}}$ is equilibrium constant for the reaction.
• $\left[{\text{BH}}^{+}\right]$ is concentration of piperazine ion.
• $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxide ion.
• $\left[\text{B}\right]$ is concentration of morphine.
• $x$ is assumed concentration of hydroxide ion.

Substitute the value of equilibrium concentration from to table to above expression.

$\begin{array}{c}{K}_{\text{b2}}=\frac{\left[{\text{BH}}_2{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{BH}}^{+}\right]}\\ 2.15\times {10}^{-9}=\frac{\left(x\right)\left(8.46\times {10}^{-4}\right)}{\left(8.46\times {10}^{-4}-x\right)}\left[8.46\times {10}^{-4}>>x\right]\\ x=2.15\times {10}^{-9}\end{array}$

Total concentration of hydroxide ion is given by,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=8.46\times {10}^{-4}+x\\ =8.46\times {10}^{-4}+2.15\times {10}^{-9}\\ =8.46002\times {10}^{-4}\text{M}\end{array}$

The $\text{pH}$ is given by the formula,

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ \text{=}14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\end{array}$

Substitute the calculated concentration of hydroxide ion in the above expression.

$\begin{array}{c}\text{pH}=14-\left(-\log \left[{\text{OH}}^{-}\right]\right)\\ =14-\left(-\log \left[8.46002\times {10}^{-4}\text{M}\right]\right)\\ =14-\left(-3.07\right)\\ =\underset{\_}{10.92}\end{array}$

Hence, the calculated $\text{pH}$ is $\underset{\_}{10.92}$.

(b)

Interpretation Introduction

To determine: The structure of piperazine ion that is present in stomach at $0.15\text{M}$

Interpretation: The structure of piperazine ion that is present in stomach is to be drawn.

Concept introduction: The $\text{pH}$ value of a compound is defined as the log of the concentration of hydronium ion released by the compound upon dissociation. The value of $\text{pH}$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $\text{pH}$ is $7$.

Answer to Problem 16.70QP

The required structure of piperazine is shown in Figure 1.

Explanation of Solution

Piperazine is basic in nature. Therefore, it releases ${\text{OH}}^{-}$ ion in its aqueous form.

The structure of the piperazine ions is,

Figure 1

Conclusion

The $\text{pH}$ of a $0.0133\text{M}$ solution of piperazine is $\underset{\_}{10.92}$.

The structure of piperazine ion is shown in Figure 1.