Question

The Ka of a monoprotic weak acid is 0.00732. What is the percent ionization of a 0.123 M solution of this acid?

Answer 1

Answer:

21.60% is the percent ionization of a 0.123 M solution of this acid.

Explanation:

The equilibrium reaction for dissociation of weak acidis,

$HA\rightleftharpoons A^-+H^+$

initially conc.         c                       0         0

At eqm.         $c(1-\alpha )$                  $c\alpha$     $c\alpha$  

Concentration of the weak acid (c) = 0.123 M

Acid dissociation constant = $k_a=0.00732$

Degree of ionization of weak acid = $\alpha$

An expression of dissociation constant is given as:

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha )}=\frac{c\times (\alpha )^2}{(1-\alpha )}$

$0.00732=\frac{0.123 M\times (\alpha )^2}{(1-\alpha )}$

$\alpha =0.2160$

Percent ionization of weak acid:

$\% ionization=\frac{c\alpha }{c}\times 100$

$\frac{0.123 M\times 0.2160}{0.123 M}\times 100=21.60\%$