Answer:
The pH in 0.140 M hippuric acid solution is 2.2.
Explanation :
Dissociation constant of the acid = [tex]K_a[/tex]
[tex]pK_a=-\log[K_a][/tex]
[tex]3.62=-\log[K_a][/tex]
[tex]K_a=2.399\times 10^{-4}[/tex]
Concentration of hippuric acid = c = 0.140 M
[tex]HC_9H_8NO_3\rightleftharpoons C_9H_8NO_{3}^-+H^+[/tex]
Initially
c 0 0
At equilibrium
(c-x) x x
Concentration of acid = c [tex] [HC_9H_8NO_3]=0.140 M[/tex]
Dissociation constant of an acid is given by:
[tex]K_a=\frac{[C_9H_8NO_{3}^-][H^+]}{[HC_9H_8NO_{3}]}[/tex]
[tex]K_a=\frac{x\times x}{(c -x)}[/tex]
[tex]2.399\times 10^{-4}=\frac{x\times x}{(0.140 -x)}[/tex]
Solving for x:
x = 0.005677 M
[tex][H^+]=x = 0.005677 M[/tex]
The pH of the solution :
[tex]pH=-\log[H^+][/tex]
[tex]pH=-\log[0.005677 M]=2.246\approx 2.2[/tex]
The pH in 0.140 M hippuric acid solution is 2.2.
