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

          NH4+ (aq) + NO2- (aq)--> HNO2 (aq) + NH3 (aq)            Given   Ka ( NH4+) = 5.6 x10-10                         Kb(NO2-)= 2.2 x 10-11               we have ,   Kb( NH3) = 1.8 x10-5                                  Ka(HNO2) = 4.6 x 10-4          Since Kb( NH3) > Ka (NH4+)  and Ka(HNO2) >Kb(NO2-) = 2.2 x 10-11 , the reverse reaction is more favourable.              HNO2(aq) + NH3(aq) -->  NH4+ (aq) +NO2- (aq)              Equilibrium constant K = Kw/  1.8 x 10-5 * 4.6 x10-4                                                        = 10-14 /  1.8 x 10-5 *4.6 x 10-4                                                          = 1.2 x 10-6               Hope this helps you.               

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