Question

Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.50 M ). However, in more concentrated solutions, part of the nitric acid is present as un-ionized molecules of HNO3. For example, in a 50% solution (7.50 M ) at 25°C, only 33% of the molecules of HNO3 dissociate into H+ and NO3–. What is the value of Ka for HNO3?

Answer 1

Answer:

The value of the $K_a$ of the nitric acid is 1.2.

Explanation:

The initial concentration of nitric acid = c = 7.50 M

$1 mM =m 10^{-3} M$

The dissociation constant of nitric acid =  $K_a$

Degree of dissociation of nitric acid = $\alpha =33\%=0.33$

$HNO_3\rightleftharpoons NO_3^{-}+H^+$

initially

c       0    0

At equilibrium

(c-cα)      cα   cα

The expression of dissociation constant :

$K_a=\frac{[NO_3^{-}][H^+]}{[HNO_3]}$

$K_a=\frac{c\times \alpha \times c\times \alpha }{c-c\alpha}$

$K_a=\frac{c\times (\alpha )^2}{(1-\alpha )}$

$K_a=\frac{7.50 M\times (0.33)^2}{(1-0.33)}=1.2$

The value of the $K_a$ of the nitric acid is 1.2.