The Ka value of nitrous acid, HNO2 , is 4.0 × 10¯4 . What is the equilibrium constant for the reaction; NO2¯ + H2O <-> HNO2 + OH¯ ?
The equation you have written is just the hydrolysis of the NO2^- so you want Kb which is Kw/Ka.
Thomas, check your 2-16-11,4:29pm post.
To find the equilibrium constant for the given reaction, we can use the equation:
Kc = [HNO2][OH¯] / [NO2¯][H2O]
Since HNO2 is the weak acid and NO2¯ is its conjugate base, we can write the equilibrium constant expression as:
Kc = [H+][NO2¯] / [HNO2][OH¯]
Now, we need to express [H+] and [OH¯] in terms of the given Ka value.
[H+] = sqrt(Ka*[HNO2]) (from the dissociation of water)
[OH¯] = Kw / [H+] = Kw / sqrt(Ka*[HNO2])
Here, Kw is the ion product of water and its value is 1.0 x 10^-14 at 25 degrees Celsius.
Substituting these expressions into the equilibrium constant expression, we have:
Kc = [H+][NO2¯] / [HNO2][OH¯]
= [sqrt(Ka*[HNO2])][NO2¯] / [HNO2][Kw / sqrt(Ka*[HNO2])]
= [NO2¯][sqrt(Ka*[HNO2]) / [HNO2]][sqrt(Ka*[HNO2]) / Kw]
= (NO2-) * ((Ka*[HNO2]) / Kw)
Finally, we substitute the given values:
Kc = (NO2-) * ((Ka*[HNO2]) / Kw)
= (1) * ((4.0 × 10^-4) * (4.0 × 10^-4) / 1.0 × 10^-14)
= 1.6 × 10^6
Therefore, the equilibrium constant (Kc) for the reaction NO2¯ + H2O <-> HNO2 + OH¯ is 1.6 × 10^6.
To find the equilibrium constant (Kc) for the reaction NO2¯ + H2O ↔ HNO2 + OH¯, you can use the relationship between the equilibrium constants of related reactions.
The given Ka value for nitrous acid, HNO2, represents the acid dissociation constant, which is defined as the ratio of the concentration of the product ions to the concentration of the undissociated acid.
HNO2 ↔ H+ + NO2¯
Ka = [H+][NO2¯] / [HNO2]
From this equation, we can see that [H+] and [NO2¯] are the concentrations of the product ions, and [HNO2] is the concentration of the undissociated acid.
Now, let's analyze the given reaction:
NO2¯ + H2O ↔ HNO2 + OH¯
We can rewrite this equation as:
[HNO2][OH¯] / [NO2¯][H2O]
The equilibrium constant (Kc) for this reaction is the ratio of the concentrations of the product species to the concentrations of the reactant species, with each concentration raised to the power of its coefficient in the balanced equation.
Thus, Kc = [HNO2][OH¯] / [NO2¯][H2O]
To relate this equilibrium constant to the given Ka value, we can make use of the autoionization of water:
H2O ↔ H+ + OH¯
The equilibrium constant for this reaction is defined as Kw, the autoionization constant of water, which has a value of 1.0 × 10^-14 at 25°C.
Using the relationship between Kc and Kw, we can write:
Kc = Kw / Ka
Substituting the given values, we have:
Kc = (1.0 × 10^-14) / (4.0 × 10^-4)
Kc = 2.5 × 10^-11
Therefore, the equilibrium constant (Kc) for the reaction NO2¯ + H2O ↔ HNO2 + OH¯ is 2.5 × 10^-11.