chemistry
posted by Dustin .
Caclulate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%
ka=(H+)(NO2)/(HNO2)
If 1.59% ionized then after ionization
H+ = 0.2 x 0.058 =?
NO2 = 0.2 X 0.058=?
HNO2=0.2 x (1.0  0.058)
If solution is 5.8% ionized, than uinionized is 100 5.8 = 94.2% or 1.000.058 = 0.942
So
ka=(0.058)(0.058)/(0.942)
=3.6 x 10^3
Why is my answer incorrect?

Caclulate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%
ka=(H+)(NO2)/(HNO2)
If 1.59% ionized then after ionization
This looks like my answer and my work from a day or so ago BUT I have no idea where the 1.59% comes from. It should say, if 5.8% is ionized, then after ionization,
H+ = 0.2 x 0.058 =? 0.0116 M.
NO2 = 0.2 X 0.058=? 0.0116 M.
HNO2=0.2 x (1.0  0.058) 0.188
If solution is 5.8% ionized, than uinionized is 100 5.8 = 94.2% or 1.000.058 = 0.942
So
ka=(0.058)(0.058)/(0.942)
Ka = (0.0116)(0.0116)/(0.188) =
7/14 x 10^4
=3.6 x 10^3
Why is my answer incorrect?
As far as I can tell you substituted the fraction ionized (for H^+, NO2^) and fraction unionized (for HNO2) instead of the concns of the H^+ and NO2^ and HNO2.Check my work carefully. 
7/14 x 10^4 should be 7.14 x 10^4. Another typo on my part.