Question

A hand book states that the solubility of ${\text{RN}}{{\text{H}}_2}{\text{ }}(g)$ in water at $1{\text{ atm}}$ and $0{{\text{ }}^ \circ }{\text{C}}$is $22.41$volumes of ${\text{RN}}{{\text{H}}_{2{\text{ }}}}(g)$ per volume of water. ( ${\text{p}}{{\text{K}}_{\text{b}}}$ of ${\text{RN}}{{\text{H}}_2}$ $ = {\text{ 4}}$ ) . Find the max. ${\text{pOH}}$ that can be obtained by dissolving ${\text{RN}}{{\text{H}}_2}$ in water:
(A)1
(B) 2
(C) 4
(D) 6

Answer 1

Hint: Here it is given that an alkyl amine is dissolved in water at a given temperature and pressure to attain a particular volume given in the question. The base dissociation constant value of the alkyl amine is given. We have to find out the maximum ${\text{pOH}}$ attained by dissolving it in water.

Complete step by step solution:
In this question, we have been provided with some information as follows:
Pressure, P = $1{\text{ atm}}$
Temperature, T = $0{{\text{ }}^ \circ }{\text{C}}$
Volume, V = $22.41$ ${\text{L}}$
${\text{p}}{{\text{K}}_{\text{b}}}$ of ${\text{RN}}{{\text{H}}_2}$ $ = {\text{ 4}}$
We also know that we can calculate the concentration of ${\text{O}}{{\text{H}}^ - }$ ions using the equation
$\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}\sqrt {{{\text{K}}_{\text{b}}} \times {\text{C}}} $
                    where, $\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ }}$is the concentration of ${\text{O}}{{\text{H}}^ - }$ ions,
                                   ${{\text{K}}_{\text{b}}}$ is the base dissociation constant
                                   And, ${\text{C}}$ is the concentration of the base.
Taking logarithm on both sides, we get,
${\text{log}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = log}}{\left( {{{\text{K}}_{\text{b}}} \times {\text{C}}} \right)^{\dfrac{1}{2}}}$
Further, we can write this equation as,
${\text{log}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}\dfrac{1}{2}{\text{log}}\left( {{{\text{K}}_{\text{b}}} \times {\text{C}}} \right)$
We can write this as,
${\text{log}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}\dfrac{1}{2}\left[ {{\text{log }}{{\text{K}}_{\text{b}}}{\text{ + log C}}} \right]$
Taking negative sign on both the sides,
${\text{ - log}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}\dfrac{1}{2}\left[ {{\text{ - log }}{{\text{K}}_{\text{b}}}{\text{ + - log C}}} \right]$
Since, we know that, ${\text{pOH = - log}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ and ${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = - log}}\left[ {{\text{p}}{{\text{K}}_{\text{b}}}} \right]$ ,
The above equation can be rewritten as,
${\text{pOH = }}\dfrac{1}{2}\left[ {{\text{p}}{{\text{K}}_{\text{b}}} - {\text{ log C}}} \right]$

Now, in order to calculate the ${\text{pOH}}$ value, we need to calculate the value of concentration of base, so that we can substitute the values in the previous equation and find out ${\text{pOH}}$ .
To calculate ${\text{C}}$ , we can use the ideal gas equation and find out the number of moles,
The ideal gas equation is,
${\text{PV = nRT}}$
Substituting the values of ${\text{P , V and T}}$ from the question and putting the value of universal gas constant as
${\text{R = 0}}{\text{.0821 L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ ,
Pressure, P = $1{\text{ atm}}$
Temperature, T = $0{{\text{ }}^ \circ }{\text{C}}$
Volume, V = $22.41$ ${\text{L}}$
Substituting in the equation,
$1{\text{ atm }} \times {\text{ 22}}{\text{. 41 L = n }} \times {\text{ 0}}{\text{.0821 L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }} \times {\text{ 273 K}}$ $\because $ $0{{\text{ }}^ \circ }{\text{C}}$ = $273{\text{ K}}$
Rearranging the unknown to the L.H.S,
${\text{n }} = {\text{ }}\dfrac{{1{\text{ atm }} \times {\text{ 22}}{\text{. 41 L}}}}{{{\text{0}}{\text{.0821 L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }} \times {\text{ 273 K}}}}{\text{ }}$
$\therefore $ ${\text{n }} \approx {\text{ 1}}$
Calculating the concentration of base, we get,
$\left[ {{\text{RN}}{{\text{H}}_2}} \right]{\text{ = }}\dfrac{{\text{n}}}{{{\text{V of solution (in L)}}}}{\text{ = }}\dfrac{{1{\text{ mol}}}}{{1{\text{ L}}}}{\text{ = 1 mol}}{{\text{L}}^{ - 1}}{\text{ = 1 M}}$

We can calculate ${\text{pOH}}$ by substituting the values of ${{\text{K}}_{\text{b}}}$ and ${\text{C}}$ in the equation,
${\text{pOH = }}\dfrac{1}{2}\left[ {{\text{p}}{{\text{K}}_{\text{b}}} - {\text{ log C}}} \right]$
We will get,
$\Rightarrow$ ${\text{pOH = }}\dfrac{1}{2}\left[ {4 - {\text{ log 1}}} \right]$
$\Rightarrow$ ${\text{pOH = }}\dfrac{1}{2}\left[ {4 - {\text{ 0}}} \right]$
$\Rightarrow$ ${\text{pOH = }}\dfrac{1}{2}\left[ 4 \right]$
$\Rightarrow$ ${\text{pOH = 2}}$
Thus, we get the value as ${\text{pOH = 2}}$ .

Hence, option (B) is the correct answer.

Note:
${{\text{K}}_{\text{a}}}$ is the acid dissociation constant of an acid where as ${{\text{K}}_{\text{b}}}$ is the base dissociation constant. ${\text{p}}{{\text{K}}_{\text{a}}}$ is the ${\text{log}}$ of acid dissociation constant and ${\text{p}}{{\text{K}}_{\text{b}}}$ is the $ - {\text{log}}$ of base dissociation constant. These constants are usually expressed in terms of ${\text{mol}}{{\text{L}}^{ - 1}}$.