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

Answer 1

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 (1-0.058) = ?? (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.

Answer 2

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 (1-0.058) = ?? (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?

Answer 3

I'm doing this exact same question lol, I got 7.1x10^-4