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Given the following equilibrium constants,

Ka (NH4^+)=5.6*10^-10
Kb (NO2^-)=2.2*10^-11
Kw=1.00*10^-14
determine the equilibrium constant for the reaction below at 25*C.

NH4^+(aq)+NO2^-(aq)f HNO2(aq)+NH3(aq)

2 answers

I assume you meant for the arrow to be as shown below:
NH4^+(aq)+NO2^-(aq)=> HNO2(aq)+NH3(aq)

Write Keq expression.
(HNO2)(NH3)/(NH4^+)(NO2^-)

Now multiply numerator and denominator by (H^)(OH^-)/(H^+)(OH^-) which is just multiplying by 1 and that doesn't change anything. Now look carefully. Note there is (HNO2)/(H^+)(NO2^-) and that is simply 1/Ka.
Note there is (NH3)/(NH4^+)(OH^-) and that is just 1/Kb.
Look to see what is left and it is (H^+)(OH^-) = Kw.
So the equilibrium constant for the reaction is Kw/KaKb. I will leave it for you to calculate the value. Note that there is NO concn of the salt which means all concns (more or less anyway) give the same pH.
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