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July 30, 2014

Homework Help: chemistry

Posted by Dustin on Monday, April 7, 2008 at 5:45pm.

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?

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