# chemistry

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

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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
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

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.

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posted by DrBob222
2. 7/14 x 10^-4 should be 7.14 x 10^-4. Another typo on my part.

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posted by DrBob222

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