Determine the percent ionization of a 0.140 m hcn solution.

                       HCN  ⇄     H⁺     +     CN⁻
Initial              0.14            0                0
Change            -x             +x             +x
Equilibrium    (0.14 -x)     +x             +x
Ka = 
6.2 x 10⁻¹⁰ = 
value of x <<<< 0.14
so x² = (6.2 x 10⁻¹⁰) (0.14)
x = 9.32 x 10⁻⁶ = [H⁺] = [CN⁻]
% ionization = Hydrogen ion concentration / total concentration x 100 
                    = 
                    = 
                    = 6.65 x 10⁻³ %

Iam not sure butch i thinnk it is ab

Sorry i dont understand the question can you rephase it?