Question

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?

Answer 1

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
<b>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,</b>
H+ = 0.2 x 0.058 =? <b> 0.0116 M.</b>
NO2- = 0.2 X 0.058=?<b> 0.0116 M. </b>
HNO2=0.2 x (1.0 - 0.058) <b>0.188</b>

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)
<b>Ka = (0.0116)(0.0116)/(0.188) =
7/14 x 10^-4 </b>

=3.6 x 10^-3

Why is my answer incorrect?
<i>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.</i>Check my work carefully.

Answer 2

7/14 x 10^-4 should be 7.14 x 10^-4. Another typo on my part.