Asked by Dustin
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?
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?
Answers
Answered by
DrBob222
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.
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.
Answered by
DrBob222
7/14 x 10^-4 should be 7.14 x 10^-4. Another typo on my part.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.