Q18.180CP - Chemistry: Molecular Nature Of Matter And Change

Question

Putrescine,$\mathbf{\left[}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}{\mathbf{\left(}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\right)}}_{\mathbf{\text{4}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}\mathbf{\right]}$ found in rotting animal tissue, is now known to be in all cells and essential for normal and abnormal (cancerous) growth. It also plays a key role in the formation of GABA, a neurotransmitter. A aqueous solution of putrescine has[ $\mathbf{\text{O}}{\mathbf{\text{H}}}^{\mathbf{\text{-}}}\mathbf{]}$=${\mathbf{\text{2.1×10}}}^{\mathbf{\text{-3}}}$. What is the${\mathbf{\text{K}}}_{\mathbf{\text{b}}}$? 

Step-by-Step Solution

Answer

_{${\text{K}}_{\text{b}}{\text{ =4.41×10}}^{\text{-5}}$}

Step 1: To find Kb

$\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{2}}\right]\text{=0.10M}\left[{\text{OH}}^{\text{-}}\right]{\text{=2.1×10}}^{\text{-3}}$ First, write the reaction equation of putescine with water.

${\text{NH}}_2{\left({\text{CH}}_2\right)}_4{\text{NH}}_2+{\text{H}}_2\text{O}\rightleftharpoons {\text{NH}}_2{\left({\text{CH}}_2\right)}_4{\text{NH}}_3^{+}+{\text{OH}}^{-}$Next, write the ${\text{K}}_{\text{b}}$expression.

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{\text{[ product ]}}{\text{[ reactant ]}}\\ {\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{3}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{2}}\right]}\end{array}$

Note that $\left[{\text{OH}}^{\text{-}}\right]\text{=}\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{3}}^{\text{+}}\right]$since they are produced 1 mole each.

$\begin{array}{c}{\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{3}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NNH}}_{\text{2}}\right]}\\ \text{=}\frac{{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}}{\left[{\text{NH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{NH}}_{\text{2}}\right]}\\ \text{=}\frac{{\left({\text{2.1×10}}^{\text{-3}}\right)}^{\text{2}}}{\text{0.10}}\\ {\text{K}}_{\text{b}}{\text{ =4.41×10}}^{\text{-5}}\end{array}$

Hence ${\text{K}}_{\text{b}}{\text{ =4.41×10}}^{\text{-5}}$