Answer:
[tex]Ka=2.99x10^{-5}[/tex]
Explanation:
Hello.
In this case, as you are not indicating the molarity of the solution and its pH we are going to assume they are 1.4 M and 2.19 respectively as typical problems over this. In such a way, based on the pH we can compute the concentration of hydrogen ions in the solution:
[tex][H^+]=10^{-pH}=10^{-2.19}=6.467x10^{-3}M[/tex]
Next, the dissociation of 4-chlorobutanoic acid is:
[tex]Cl-CH_2-CH_2-CH_2-COOH\rightleftharpoons H^++Cl-CH_2-CH_2-CH_2-COO^-[/tex]
Since it is a weak acid, therefore, its equilibrium expression is:
[tex]Ka=\frac{[H^+][Cl-CH_2-CH_2-CH_2-COO^-]}{[Cl-CH_2-CH_2-CH_2-COOH]}[/tex]
Since the concentration of hydrogen ions equal the concentration of 4-chlorobutanoate ions at equilibrium and the concentration the 4-chlorobutanoic acid equals 1.4 M minus the concentration of hydrogen ions, which is related to the reaction extent [tex]x[/tex], the acid dissociation constant is:
[tex]Ka=\frac{6.467x10^{-3}*6.467x10^{-3}}{1.4-6.467x10^{-3}}\\ \\Ka=2.99x10^{-5}[/tex]
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