Question

What is the percent ionization of 0.45M hydrocyanic acid (HCN, pKa =9.21)

Answer 1

Dear Student,

Given:

Molarity of HCN = 0.45 M

pKa = 9.21

To calculate percent ionization

Formula:

pKa = ( c x c) / c(1-)

Solution

HCN            H⁺    +   CN^-

c(1-)              c         c

pKa = [H⁺ ] [CN^- ] / [HCN]

pKa = c x c / c(1-)             

9.21 = c²² / c (1-)

9.21 = c ² / 1

Here (1-) is approx to 1

9.21 = 0.45 x ²

² = 9.21 / 0.45

² = 20.46

= 20.46

= 4.52 M

Therefore,the concentration of H⁺ is 4.52 M.

Now, divide the [H⁺] by the given concentration and then multiply by 100 we get,

Percent Ionization =( [H⁺] / 0.45 ) x 100

Percent Ionization = (4.52 / 0.45 ) x 100

Percent Ionization = 10.04 x 100

Percent Ionization = 1004 %

Therefore, the percent ionization of 0.45 M hydrocyanic acid is 1004%.