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 1.00-0.058 = 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 1.00-0.058 = 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.

There seems to be a minor error in your calculation. Let's correct it step-by-step:

Given:
Percent ionization = 5.8%
[H+] = percent ionization × initial concentration = 0.058 × 0.200 = 0.0116 mol/L
[NO2-] = percent ionization × initial concentration = 0.058 × 0.200 = 0.0116 mol/L
[HNO2] = initial concentration - [H+] = 0.200 - 0.0116 = 0.1884 mol/L

Now, substitute these values into the Ka expression:
Ka = ([H+][NO2-]) / [HNO2]
Ka = (0.0116)(0.0116) / 0.1884
Ka ≈ 0.000072

So, the correct value for the Ka of nitrous acid is approximately 0.000072.

Your answer is incorrect because you made a mistake in calculating the concentrations of the ions. Let me walk you through the correct steps:

Given that the solution is 5.8% ionized, it means that 5.8% of the initial concentration of nitrous acid (HNO2) has dissociated into its ions, H+ and NO2-.

To calculate the concentration of H+, you multiply the initial concentration of the acid (0.200 mol/L) by the percent ionization (5.8% or 0.058):

H+ = 0.200 mol/L x 0.058 = 0.0116 mol/L

To calculate the concentration of NO2-, you also multiply the initial concentration of the acid (0.200 mol/L) by the percent ionization (0.058):

NO2- = 0.200 mol/L x 0.058 = 0.0116 mol/L

To calculate the concentration of the remaining undissociated acid (HNO2), you subtract the concentration of the H+ ions from the initial concentration of the acid:

HNO2 = 0.200 mol/L - 0.0116 mol/L = 0.1884 mol/L

Now, using these correct concentrations, we can calculate the Ka of nitrous acid:

Ka = (H+)(NO2-) / (HNO2)
= (0.0116 mol/L)(0.0116 mol/L) / (0.1884 mol/L)
≈ 7.15 x 10^-4

So, the correct Ka value for nitrous acid is approximately 7.15 x 10^-4.