Answer:
[tex]Ka=1.11x10^{-3}[/tex]
Explanation:
Hello,
In this case, since the ionization of the given HA acid is:
[tex]HA\rightleftharpoons H^++A^-[/tex]
The equilibrium expression is:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Whereas the concentration of hydrogen ions is compute from the pH=
[tex][H^+]=10^{-pH}=10^{-2.00}=0.01M[/tex]
Which also equals the concentration of [tex]A^-[/tex] and the in general the ionization extent, therefore, the acid ionization constant, Ka, turns out:
[tex]Ka=\frac{0.01*0.01}{0.1-0.01}\\ \\Ka=1.11x10^{-3}[/tex]
Regards.
