Posted by Laura on .
Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%

chemistry 
DBob222,
HNO2 ==> H^+ + NO2^
Initial:
(HNO2) = 0.2
(H^+) = 0
(NO2^) = 0
Ka = (H^+)(NO2^)/(HNO2)
If 5.8% ionized, then after ionization:
(H^+) = 0.2 x 0.058 = ??
(NO2^) = 0.2 x 0.058 = ??
(HNO2) = 0.2 x (10.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100  5.8 = 94.2% OR 1.00  0.058 = 0.942).
Plug these values into Ka expression above and solve for Ka. 
chemistry 
Sarah,
Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%
For Further Reading
chemistry  DBob222, Saturday, April 5, 2008 at 4:50pm
HNO2 ==> H^+ + NO2^
Initial:
(HNO2) = 0.2
(H^+) = 0
(NO2^) = 0
Ka = (H^+)(NO2^)/(HNO2)
If 5.8% ionized, then after ionization:
(H^+) = 0.2 x 0.058 = ??
(NO2^) = 0.2 x 0.058 = ??
(HNO2) = 0.2 x (10.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100  5.8 = 94.2% OR 1.00  0.058 = 0.942).
Plug these values into Ka expression above and solve for Ka.
Here is my work:
Ka = (H^+)(NO2^)/(HNO2)
= (0.0116)(0.0116) / (0.1884)
= 1.34 X 10^4 / (0.1884)
= 7.14 X 10^4
My answer is still wrong:( Is there a step I missed?