The given reaction is
Now, we are given that the initial concentration of the conjugate base NO2- is 0.69 M, which results in the given ICE table
Now, we can write the expression of equilibrium constant K for the above reaction as follows:
Where the concentrations are the equilibrium values.
Note that H2O does not appear in the expression as it is the pure solvent with a constant concentration.
Since we have a base (NO2-) being hydrolysed in water, the equilibrium constant is called Kb.
Now, using the ICE table we can write the following expression for Kb.
Now, it is also given that the Ka of HNO2 is as .
The reaction of HNO2 in water for which we get the equilibrium constant Ka can be written as
Hence, we can write the expression of Ka as follows:
Again water does not appear in the expression.
Now if we multiply expressions of Ka and Kb we get the following,
The product is called the ionic product of water, Kw and has a constant value for an aqueous solution at a particular temperature.
At standard temperature of 25 C,
Hence, we can write
Hence, the Kb of the reaction given in the question is approximately (rounded to two significant digits.)
Now, using the Kb value calculated above, we can calculate the equilibrium concentrations of the species using the ICE table as follows:
Hence, the equilibrium concentrations in the ICE table will be
please dive details on how you got this answer also...thanks! + H2O() $ NO- (aq) 0.69M...
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