Respuesta :
Answer : The pH of the solution is, 3
Solution : Given,
Concentration (C) = 0.06 M
Acid dissociation constant = [tex]k_a=1.78\times 10^{-5}[/tex]
The equilibrium reaction for dissociation of [tex]CH_3COOH[/tex] (weak acid) is,
[tex]CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+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 [tex][H^+][/tex]
As, we know that
[tex]\alpha=\sqrt{\frac{k_a}{c}}[/tex] (for weak electrolyte) ...........(1)
where, [tex]\alpha[/tex] is degree of dissociation
[tex][H^+]=c\alpha[/tex] ..................(2)
By equation both the equations (1) and (2), we get
[tex][H^+]=\sqrt{k_a\times c}[/tex]
Now put all the given values in this expression, we get
[tex][H^+]=\sqrt{1.78\times 10^{-5}\times 0.06}=1.033\times 10^{-3}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (1.033\times 10^{-3})[/tex]
[tex]pH=2.8=3[/tex]
Therefore, the pH of the solution is, 3
