Respuesta :
The dissociation reaction of acetic acid is as follows:
[tex]CH_{3}COOH\leftrightharpoons CH_{3}COO^{-}+H^{+}[/tex]
The acid dissociation constant K_{a} is [tex]1.76\times 10^{-5}[/tex].
Let the initial concentration of acid be A, and concentration of [tex]CH_{3}COO^{-}[/tex] and [tex]H^{+}[/tex] be zero.
After dissociation, concentration of acid becomes A-x and that of both [tex]CH_{3}COO^{-}[/tex] and [tex]H^{+}[/tex] becomes x.
Expression for acid dissociation constant will be:
[tex]K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}[/tex]
pH of solution is 3.5, thus, concentration of hydrogen ion can be calculated as follows:
[tex]pH=-log[H^{+}][/tex]
On rearranging,
[tex][H^{+}]=10^{-pH}=10^{-3.5}=0.0003162[/tex]
Since, [tex][CH_{3}COO^{-}]=[H^{+}]=x[/tex]
Thus,
[tex][CH_{3}COO^{-}]=0.0003162[/tex]
and, [tex]x=0.0003162[/tex]
Putting the values, in expression for acid dissociation constant,
[tex]1.76\times 10^{-5}=\frac{(0.0003162)(0.0003162)}{[CH_{3}COOH]_{initial}-0.0003162}[/tex]
On rearranging,
[tex][CH_{3}COOH]_{initial}=\frac{(0.0003162)\times (0.0003162)}{1.76\times 10^{-5}}+0.0003162=0.006[/tex]
