Posted by Dustin on Monday, April 7, 2008 at 5:45pm.
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 1.00-0.058 = 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.
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