Given the following equilibrium constants, Ka (NH 4 + )= 5.6E-10 Kb (NO 2 - ) =

Question

Given the following equilibrium constants, Ka (NH4+)= 5.6E-10 Kb (NO2-) = 2.2E-11 Kw = 1.0E-14 determine the equilibrium constant for the reaction belowat 25degrees C NH4+ (aq) + NO2-(aq) --> HNO2 (aq) + NH3 (aq) the answer is 1.2E-6 but i don't know how thanks Given the following equilibrium constants, Ka (NH4+)= 5.6E-10 Kb (NO2-) = 2.2E-11 Kw = 1.0E-14 determine the equilibrium constant for the reaction belowat 25degrees C NH4+ (aq) + NO2-(aq) --> HNO2 (aq) + NH3 (aq) the answer is 1.2E-6 but i don't know how thanks

Explanation / Answer

We Know that :        NH4OH (aq.) <------> NH4+ (aq) + OH- (aq.)         HNO2 (aq.) <------> H+ (aq.) + NO2- (aq.)           H2O (l) <-------> H+ (aq.)   + OH - (aq.)          The givenequation is :           NH4+ (aq) + NO2- (aq)--> HNO2 (aq) + NH3 (aq)          K = [ HNO2 ] [ NH3] / [NH4+ ] [ NO2-]               = Ka x Kb / Kw                = 5.6 x 10-10 x 2.2 x10-11   / 1.0 x 10-14                 = 1.2 x 10-6                 = 1.2 x 10-6

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