Question

please dive details on how you got this answer also...thanks!

+ H2O() $ NO- (aq) 0.69M HNO2 (ag) + OH(aq) -1x +1x + 1x 0.69-1x +1x +1x Ε The K for HNO, is 4.6 x 104. What is the value of Ko for the reaction above?

Answer 1

The given reaction is

NO2 (aq) + H2O) = HNO2(aq) + OH

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:

K_(HNO3)(OH) [NO]

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 K_b.

Now, using the ICE table we can write the following expression for K_b.

K. (HNO2](OH- [NO] 1x x 1. (0.69 – 1.c)

Now, it is also given that the K_a of HNO₂ is as .

4.6 x 10-4

The reaction of HNO₂ in water for which we get the equilibrium constant K_a can be written as

HNO3(aq) + H2O) = H30ag) + NO2 (aq)

Hence, we can write the expression of K_a as follows:

K [NO] [H30+1 [H NO2]

Again water does not appear in the expression.

Now if we multiply expressions of K_a and K_b we get the following,

Kax Ko = [107][H30+1, [HNO2][OH-] T = [H30+][OH [HNO2 NO,

The product is called the ionic product of water, K_w and has a constant value for an aqueous solution at a particular temperature.

Holl+oʻH

At standard temperature of 25 C,

Kw = [H3O+1OH] = 1.0 x 10-14

Hence, we can write

Kw = K x K = 1.0 X 10-14 →K- Kw_1.0 x 10-14 K 4.6 x 10-4 h =K 4 2.174 x 10-11 .6 x 10

Hence, the K_b of the reaction given in the question is approximately
2.2 x 10-11
(rounded to two significant digits.)

Now, using the K_b value calculated above, we can calculate the equilibrium concentrations of the species using the ICE table as follows:

K6 = [H NO2][OH-] 4 1.2 x 12 [NO] (0.69 – 1.0) -= 2.174 x 10-11 → 32 = 2.174 x 10-11 (0.69 – ) 3,87 x 10-6 M

Hence, the equilibrium concentrations in the ICE table will be

HNO2) = [OH-] = 1.1 = 3.87 x 10-6 M

N07] = 0.69 M - I = 0.69 M - 3.87 x 10-6 M -0.69 M