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)
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.
76
To determine the equilibrium constant (K) for the given reaction, we can use the equation:
K = (Kc1 * Kc2) / (Kc3 * Kc4)
Where Kc1, Kc2, Kc3, and Kc4 are the equilibrium constants for the individual reactions involved in the overall reaction.
The overall reaction can be divided into two steps:
Step 1: NH4^+(aq) + OH^-(aq) -> NH3(aq) + H2O(l)
Step 2: HNO2(aq) + OH^-(aq) -> NO2^-(aq) + H2O(l)
Step 1 involves the ionization of ammonia (NH3) and water (H2O). The equilibrium constant for this reaction can be expressed as Kw, the ion product of water, which is given as 1.00 * 10^-14.
Step 2 involves the ionization of nitrous acid (HNO2) and water (H2O). The equilibrium constant for this reaction is Kb(NO2^-), given as 2.2 * 10^-11.
Now, let's calculate K for the overall reaction:
K = (Kc1 * Kc2) / (Kc3 * Kc4)
K = (Kw * Kb(NO2^-)) / (Ka(NH4^+) * 1)
Substituting the given values:
K = (1.00 * 10^-14) * (2.2 * 10^-11) / (5.6 * 10^-10)
K ≈ 3.18 * 10^-15
Therefore, the equilibrium constant for the given reaction at 25°C is approximately 3.18 * 10^-15.