Respuesta :
Answer:
2.872.
Explanation:
- For weak acids: [H⁺] = √(Ka.C).
∴ [H⁺] = √(Ka.C) = √(1.8 x 10⁻⁵)(0.1 M) = 1.342 x 10⁻³.
∵ pH = - log[H⁺].
∴ pH = - log(1.342 x 10⁻³) = 2.872.
Answer : The pH of the solution is, 2.87
Solution : Given,
Concentration (c) = 0.100 M
Acid dissociation constant = [tex]k_a=1.8\times 10^{-5}[/tex]
The equilibrium reaction for dissociation of [tex]CH_3COOH[/tex] is,
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/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]1.8\times 10^{-5}=\frac{(0.1\alpha)(0.1\alpha)}{0.1(1-\alpha)}[/tex]
By solving the terms, we get
[tex]\alpha=0.0133[/tex]
Now we have to calculate the concentration of hydrogen ion.
[tex][H^+]=c\alpha=0.1\times 0.0133=1.33\times 10^{-3}M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (1.33\times 10^{-3})[/tex]
[tex]pH=2.87[/tex]
Therefore, the pH of the solution is, 2.87
