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Alex73 [517]
3 years ago
12

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

Chemistry
1 answer:
MaRussiya [10]3 years ago
5 0

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\%

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