Question

Calculate the degree of ionization of 0.01M solution of HCN.Ka = 4.8 X 10-10.Also calculate H+ ion concentration of the solution

Answer 1

Explanation:

As HCN is a weak acid as evidenced by low Ka value, α value is negligible and 1-α = 1 and 0.01(1-α) = 0.01

Ka= [H3O+] [CN−][HCN]= (0.01α) (0.01α)0.014.8 × 10−10 = 0.01 α2α2 = (4.8 × 10−10)(0.01) = 4.8 ×10−12α =4.8 ×10−12−−−−−−−−−−√ = 2.2 ×10−6

degree of dissociation is 2.2 X 10-6

Answer 2

The degree of ionization of 0.01M solution of HCN is $2.2 X 10^{-6}$

Explanation:

• The degree of ionization is expressed as the strength of ionizability of an electrolyte.It is denoted by α
• It also refers to the strength of an acid or base in order to form ions in the solution.
• In Strong acids and base complete ionization takes place.
• Whereas in weak acid and base partial ionization takes place.

Here the $HCN$ is a weak acid that has a low value of $k_{a}$

$HCN ---- H^{+}+ CN^{-}$

$\alpha$ value is negligible and $1-\alpha =1$  and  $0.01(1-\alpha )=0.01$

The value of $Ka= [H3O+] [CN^{-} ][HCN]$

$= (0.01\alpha ) (0.01\alpha )0.014.8 *10^{-10}$

$=2.2$×$10^{-6}$

Degree of dissociation of HCN is $=2.2$×$10^{-6}$