Question

The pH of 0.005 M codeine ($C_{18}H_{21}NO_{3}$) solution is 9.95. Calculate its ionization constant and pKb.

Answer 1

From the given,

Concentration of hydronium ion = 0.005 M

pH = 9.95

pH + pOH = 14

9.95 + pOH = 14

pOH = 4.05

pH = -log(4.05)

4.05 = -log[${ OH }^{ - }$]

           $[{ OH }^{ - }]\quad =\quad 8.91\quad \times \quad { 10 }^{ -5 }$

           $\alpha \quad =\quad \frac { 8.91\quad \times \quad { 10 }^{ -5 } }{ 5\quad \times \quad { 10 }^{ -3 } }$

           $\alpha \quad =\quad 1.782\quad \times \quad { 10 }^{ -2 }$

Therefore, ${ K }_{ b }\quad =\quad C{ \alpha}^{ 2 }$

           $=\quad 0.005\quad \times \quad { (1.782) }^{ 2 }\quad \times \quad { 10 }^{ -4 }$

           $=\quad 0.005\quad \times \quad 3.1755\quad \times \quad { 10 }^{ -4 }$

           ${ K }_{ b }\quad =\quad 0.015\quad \times \quad { 10 }^{ -4 }$

Therefore, the ionization constant is $0.015\quad \times\quad { 10 }^{ -4 }$

           ${ K }_{ b }\quad =\quad 1.58\quad \times \quad { 10 }^{ -6 }$

           $p{ K }_{ b }\quad =\quad -log{ K }_{ b }$

           $=\quad -log(1.58\quad \times \quad { 10 }^{ -6 })\quad =\quad 5.80$

Therefore, $p{ K }_{ b }$ is 5.80

Answer 2

HEY MATE.....
pKb =5.8_______Answer☺✌☺