Answer:
The value of the [tex]K_a[/tex] of the nitric acid is 1.2.
Explanation:
The initial concentration of nitric acid = c = 7.50 M
[tex]1 mM =m 10^{-3} M[/tex]
The dissociation constant of nitric acid = Â [tex] K_a[/tex]
Degree of dissociation of nitric acid = [tex]\alpha =33\%=0.33[/tex]
[tex]HNO_3\rightleftharpoons NO_3^{-}+H^+[/tex]
initially
c    0   0
At equilibrium
(c-cα)    cα  cα
The expression of dissociation constant :
[tex]K_a=\frac{[NO_3^{-}][H^+]}{[HNO_3]}[/tex]
[tex]K_a=\frac{c\times \alpha \times c\times \alpha }{c-c\alpha}[/tex]
[tex]K_a=\frac{c\times (\alpha )^2}{(1-\alpha )}[/tex]
[tex]K_a=\frac{7.50 M\times (0.33)^2}{(1-0.33)}=1.2[/tex]
The value of the [tex]K_a[/tex] of the nitric acid is 1.2.