Question

An unknown weak base with a concentration of 0.0910 M has a pH of 10.50. What is the Kb of this base

Answer 1

Answer: The $K_b$ for the weak base is $3.5\times 10^{-3}$

Explanation:

$A^-+H_2O\rightarrow HA+OH^-$

cM                   0          0

$c-c\alpha$           $c\alpha$       $c\alpha$ 

So dissociation constant will be:

$K_b=\frac{[HA][OH^-]}{[A^-]}$

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given :

c= 0.0910 M and pH = 10.50

pOH = 14-pH = 14-10.50 = 3.5

Also $pOH=-log[OH^-]$

$[OH^-]=antilog(-3.5)= 3.2\times 10^{-4}M$

$[OH^-]=c\alpha=3.2\times 10^{-4}$

$K_b=\frac{(3.2\times 10^{-4})^2}{(0.0910-3.2\times 10^{-4}}$

$K_b=\frac{(3.2\times 10^{-4})^2}{(0.09068}$

$K_b=3.5\times 10^{-3}$

Thus $K_b$ for the weak base is $3.5\times 10^{-3}$