Calculate the equilibrium concentration of OH-, HNO2, NO2-, and H3O+, and the ph at 25 degree...

Calculate the equilibrium concentration of OH-, HNO2, NO2-, and H3O+, and the ph at 25 degree C of a solution that is 0.25 M in NaNO2? Comment

Expert Solution

HNO2 , Ka = 5.6 x 10^-4

Kb = Kw / Ka = 1.0 x 10^-14 / 5.6 x 10^-4

Kb = 1.79 x 10^-11

NO2 - + H2O -----------------------------> HNO2 + OH-

0.25 0 0 --------------> initial

0.25-x x x ------------------> equilibrium

Kb = [HNO2][OH-]/[NO2-]

Kb = x^2 / 0.25-x

1.79 x 10^-11 = x^2 / 0.25-x

x^2 + 1.79 x 10^-11 x - 4.46 x 10^-12 =0

x = 2.11 x 10^-6

[OH-] = [HNO2] = x

[OH-] = [HNO2] = 2.11 x 10^-6 M

[NO2-] = 0.25 -x = 0.2499 = 0.25 M

[H3O+] = Kw / [OH-] = 1.0 x 10^-14 / 2.11 x 10^-6

[H3O+] = 4.74 x 10^-9 M

queen_honey_blossom answered 2 years ago

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