Question

What is the percent dissociation of HNO2 when 0.082 g of sodium nitrite is added to 115.0 mL of a 0.071 M HNO2 solution? Ka for HNO2 is 4.0 × 10–4.

Answer 1

HNO2 moles = M x V = 0.071 x 115/1000 = 0.008165

NaNO2 moles= mass/molar mass = 0.082/68.9953 = 0.00119

let X be amount dissociated on HNO2

HNO2 <---> H+ (aq) + NO2- (aq)

HNO2 moles = 0.008165-x , NaNO2 moles = 0.00119+X , [H+] moles = X

Ka = [NaNO2] [H+]/[HNO2]    , conc = moles/vol where vol = 0.115L

4x10^-4 = ( 0.00119+X) (X) / ( 0.115^2) / ( 0.008165-X) /0.115

0.46 x 10^-4 = ( X^ 2+ 0.00119X) / ( 0.008165-X)

X^2 + 0.00119X = 3.756 x 10^-7 -0.46x10^-4X

X = 0.00025236 = H+ moles

% dissociation = 100 x ( H+ moles/HNO2 moles initially) = ( 100 x 0.00025236 / 0.008165)

              = 3.09 %