Answer: [tex]1.67\times 10^{-3}[/tex]
Explanation:
[tex]ClCH_2COOH\rightarrow ClCH_2COO^-+H^+[/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]
Given: c = 0.15 M
pH = 1.86
[tex]K_a[/tex] = ?
Putting in the values we get:
Also [tex]pH=-log[H^+][/tex]
[tex]1.86=-log[H^+][/tex]
[tex][H^+]=0.01[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]0.01=0.15\times \alpha[/tex]
[tex]\alpha=0.06[/tex]
As [tex][H^+]=[ClCH_2COO^-]=0.01[/tex]
[tex]K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}[/tex]
[tex]K_a=1.67\times 10^{-3][/tex]
Thus the vale of [tex]K_a[/tex] for the acid is [tex]1.67\times 10^{-3}[/tex]
