Question

Be sure to answer all parts.

Determine the pH and percent ionization for a hydrocyanic acid (HCN) solution of concentration 3.6 ×10⁻³M (K_a for HCN is 4.9 ×10⁻¹⁰.)

pH =      

percent ionization = %
(Enter your answer in scientific notation.)

Answer 1

1)

HCN dissociates as:

HCN -----> H+ + CN-

3.6*10^-3 0 0

3.6*10^-3-x x x

Ka = [H+][CN-]/[HCN]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.9*10^-10)*3.6*10^-3) = 1.328*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.328*10^-6 M

so.[H+] = x = 1.328*10^-6 M

use:

pH = -log [H+]

= -log (1.328*10^-6)

= 5.88

Answer: pH = 5.88

B)

% dissociation = (x*100)/c

= 1.328*10^-6*100/0.0036

= 0.0369 %

Answer: 0.0369 %