Respuesta :
Answer : The pH of the solution is, 4.25
Solution : Given,
Concentration (C) = [tex]3.010\times 10^{-4}M[/tex]
Acid dissociation constant = [tex]k_a=1.310\times 10^{-5}[/tex]
The equilibrium reaction for dissociation of [tex]C_2H_5CO_2H[/tex] (weak acid) is,
[tex]C_2H_5CO_2H\rightleftharpoons C_2H_5CO_2^-+H^+[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
where, [tex]\alpha[/tex] is degree of dissociation
First we have to calculate the value of [tex]\alpha[/tex]
[tex]K_a=\frac{(c\alpha)\times (c\alpha)}{c(1-\alpha)}[/tex]
[tex]K_a=\frac{c(\alpha)^2}{(1-\alpha)}[/tex]
Now put all the given values in this formula, we get:
[tex]1.310\times 10^{-5}=\frac{(3.010\times 10^{-4})(\alpha)^2}{(1-\alpha)}[/tex]
[tex]\alpha=0.188[/tex]
Now we have to calculate the hydrogen ion concentration.
[tex][H^+]=c\alpha=(3.010\times 10^{-4})\times (0.188)=5.66\times 10^{-5}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (5.66\times 10^{-5})[/tex]
[tex]pH=4.25[/tex]
Therefore, the pH of the solution is, 4.25
