Answer :
Answer: The pH of an aqueous solution of .25M acetic acid is 2.7
Explanation:
[tex]HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.25 M and [tex]\alpha[/tex] = ?
[tex]K_a=1.8\times 10^{-5}[/tex]
Putting in the values we get:
[tex]1.8\times 10^{-5}=\frac{(0.25\times \alpha)^2}{(0.25-0.25\times \alpha)}[/tex]
[tex](\alpha)=0.0084[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.25\times 0.0084=0.0021[/tex]
Also [tex]pH=-log[H^+][/tex]
[tex]pH=-log[0.0021]=2.7[/tex]
Thus pH is 2.7