Respuesta :
Answer: The percent ionization is higher for 0.010 M solution of [tex]HC_2H_3O_2[/tex]
Explanation:
[tex]HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-[/tex]
cM 0M 0M
[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]
a) Given c= 0.10 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{(0.10\times \alpha)^2}{(0.10-0.10\times \alpha)}[/tex]
[tex](\alpha)=0.013[/tex]
[tex]\%(\alpha)=0.013\times 100=1.3\%[/tex]
b) Given c= 0.010 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{(0.010\times \alpha)^2}{(0.010-0.010\times \alpha)}[/tex]
[tex](\alpha)=0.041[/tex]
[tex]\%(\alpha)=0.041\times 100=4.1\%[/tex]
The percent ionization is higher for 0.010 M solution of [tex]HC_2H_3O_2[/tex]
