Question

Determine the percent ionization of a 0.014 M solution of hypochlorous acid, HClO. The Ka for the acid is 3.5x10^–8.

A) 3.5x10 ^–6 %

B) 4.9x10^–9 %

C) 7.0 x 10^–3 %

D) 5.0 x10^–2 %

E) 1.58x 10^–1 %

Answer 1

HClO----> H+ +ClO-

Initial                              HClO     0.014     H+   0     ClO-     0

let x= drop in concentration of HClO due to equilibrium

Equilibrium                    HClO      0.014-x      H+    x    ClO-    x

Ka= [H+] [ClO-]/ [HClO] =3.5*10^-8

x²/(0.014-x)= 3.5*10^-8,

Looking at the magnitude to K, it is reasonable to approximate 0.014-x as 0.014

x2/0.014= 3.5*10^-8

x2= 3.5*0.014*10^-8

x =0.000021

percent ionization=100*0.000021/0.014=0.15% ( E is the closest answer)