Question

The pH of $0 \cdot 05{\text{ M}}$ aqueous solution of diethyl amine is ${\text{12}}$. Calculate its ${K_{\text{b}}}$.

Answer 1

Hint: The measure basicity or the strength of base is known as base dissociation constant $\left( {{K_{\text{b}}}} \right)$. Calculate the pOH from the pH given which gives the concentration of ${\text{O}}{{\text{H}}^ - }$.
Diethyl amine dissociates as shown in the reaction, ${\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}}_2^ + + {\text{O}}{{\text{H}}^ - }$. Setup the equilibrium table and calculate the base dissociation constant.

Complete step by step answer:
Step 1:
Calculate the pOH using the equation as follows:
${\text{pH}} + {\text{pOH}} = 14$
Rearrange the equation for pOH as follows:
${\text{pOH}} = 14 - {\text{pH}}$
Substitute ${\text{12}}$ for pH. Thus,
${\text{pOH}} = 14 - 12 = 2$
Thus, the pOH is $2$.
Step 2:
Calculate the concentration of ${\text{O}}{{\text{H}}^ - }$ using the equation as follows:
${\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]$
Rearrange the equation for the concentration of ${\text{O}}{{\text{H}}^ - }$ as follows:
$\left[ {{\text{O}}{{\text{H}}^ - }} \right] = {10^{ - {\text{pOH}}}}$
Thus,
$\left[ {{\text{O}}{{\text{H}}^ - }} \right] = {10^{ - 2}}{\text{ M}}$
Thus, the concentration of ${\text{O}}{{\text{H}}^ - }$ is ${10^{ - 2}}{\text{ M}}$.
Step 3:
Calculate the base dissociation constant as follows:
The dissociation of diethylamine occurs as follows:
${\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}}_2^ + + {\text{O}}{{\text{H}}^ - }$
At equilibrium:
${\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}}_2^ + + {\text{O}}{{\text{H}}^ - }$
${\text{ 0}} \cdot {\text{05 0 0}}$
${\text{ 0}} \cdot {\text{05-x x x}}$

Thus, $x = \left[ {{\text{O}}{{\text{H}}^ - }} \right] = {10^{ - 2}}{\text{ M}} = 0 \cdot 01{\text{ M}}$
Calculate the base dissociation constant as follows:
${K_{\text{b}}} = \dfrac{{\left[ {{{\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)}_{\text{2}}}{\text{NH}}_2^ + } \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)}_{\text{2}}}{\text{NH}}} \right]}}$
${K_{\text{b}}} = \dfrac{{\left( x \right)\left( x \right)}}{{\left( {0 \cdot 05 - x} \right)}}$
${K_{\text{b}}} = \dfrac{{\left( {0 \cdot 01} \right)\left( {0 \cdot 01} \right)}}{{\left( {0 \cdot 05 - 0 \cdot 01} \right)}}$
${K_{\text{b}}} = \dfrac{{{{\left( {0 \cdot 01} \right)}^2}}}{{0 \cdot 04}}$
${K_{\text{b}}} = 2 \cdot 5 \times {10^{ - 3}}$
Thus, the base dissociation constant is $2 \cdot 5 \times {10^{ - 3}}$.

Note:
Diethyl amine dissociates as shown in the reaction, ${\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)_{\text{2}}}{\text{NH}}_2^ + + {\text{O}}{{\text{H}}^ - }$. Setup the equilibrium table and calculate the base dissociation constant.