Respuesta :
Answer : The pH of the solution is, 2.03
Solution : Given,
Concentration (c) = 0.20 M
Acid dissociation constant = [tex]k_a=4.6\times 10^{-4}[/tex]
The equilibrium reaction for dissociation of [tex]HNO_2[/tex] (weak acid) is,
[tex]HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
First we have to calculate the concentration of value of dissociation constant [tex](\alpha}[/tex].
Formula used :
[tex]k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}[/tex]
Now put all the given values in this formula ,we get the value of dissociation constant [tex](\alpha}[/tex].
[tex]4.6\times 10^{-4}=\frac{(0.2\alpha)(0.2\alpha)}{0.2(1-\alpha)}[/tex]
By solving the terms, we get
[tex]\alpha=0.0468[/tex]
No we have to calculate the concentration of hydronium ion or hydrogen ion.
[tex][H^+]=c\alpha=0.2\times 0.0468=9.36\times 10^{-3}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (9.36\times 10^{-3})[/tex]
[tex]pH=2.03[/tex]
Therefore, the pH of the solution is, 2.03
