Question

The $pH$ of 0.005 M codeine $({C_{18}}{H_{21}}N{O_3})$ solution is 9.95. Calculate its ionization constant and $p{K_b}$.

Answer 1

We have $pH$ , value using that we can calculate hydrogen ion concentration. Knowing hydrogen ion concentration and ionic product of water can calculate the concentration of hydroxyl ions. Then we can find ionization constant ${K_b}$ by using ${K_b} = \dfrac{{[{M^ + }][O{H^ - }]}}{{[MOH]}}$ , where $[{M^ + }]$ is codeine ion. We can find $p{K_b}$ by $p{K_b} = - \log ({k_b})$ .

Complete step by step solution:
We have,
$Codeine + {H_2}O\underset {} \leftrightarrows Codeine{H^ + } + O{H^ - }$
$pH = 9.95$ and $[MOH] = 0.005M$
$[{H^ + }] = anti\log \left( { - pH} \right)$
Substituting we have,
$[{H^ + }] = anti\log \left( { - 9.95} \right)$
$[{H^ + }] = 1.12 \times {10^{ - 10}}M$ .
So,
$[O{H^ - }] = \dfrac{{{K_w}}}{{[{H^ + }]}}$
${K_w}$ is an ionization constant of water. That is ${K_w} = 1 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}$ .
Substituting we have
$[O{H^ - }] = \dfrac{{1 \times {{10}^{ - 14}}}}{{1.12 \times {{10}^{ - 10}}}}$
$[O{H^ - }] = 8.928 \times {10^{ - 5}}M$
The concentration of the corresponding codenium ion is also the same as that of hydroxyl ion. Let the codeine ion be denoted by $[{M^ + }]$ . Thus
${K_b} = \dfrac{{[{M^ + }][O{H^ - }]}}{{[MOH]}}$
Substituting we have,
${K_b} = \dfrac{{{{\left( {8.928 \times {{10}^{ - 5}}} \right)}^2}}}{{0.005}}$
${K_b} = \dfrac{{7.971 \times {{10}^{ - 9}}}}{{0.005}}$
$\Rightarrow {K_b} = 1.594 \times {10^{ - 6}}$
The ionization constant is $1.594 \times {10^{ - 6}}$ .
Now $p{K_b} = - \log ({k_b})$
$p{K_b} = - \log (1.594 \times {10^{ - 6}})$
$\Rightarrow p{K_b} = 5.79$

Note:
We know that there is no unit for ionization constant and $p{K_b}$ . Careful in the calculation part. the ionization for a general weak acid, HA can be written as follows
$H{A_{(aq)}} \to {H^ + }_{(aq)} + {A^ - }_{(aq)}$
The Acid Ionization constant ${K_a}$ is given by ${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$ . Also codeine is a weak organic base.