Answer:- [tex]Ka=9*10^-^5[/tex]
Solution:- percent ionization = [tex](\frac{x}{c})100[/tex]
where, [tex]x[/tex] is the equilibrium concentration of product ion and c is the initial concentration of the acid.
Let's plug in the values in the formula:
[tex]3.0=(\frac{x}{0.10})100[/tex]
[tex]x=\frac{3.0*0.10}{100}[/tex]
[tex]x=0.003[/tex]
In general, the acid is represented as HA, the ice table is shown as:
[tex]HA(aq) \leftrightharpoons H^+(aq) +A^-(aq)[/tex]
I 0.10 0 0
C -X +X +X
E (0.10 - X) X X
where X is the change in concentration that we already have find out using percentage ionization formula.
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Let's plug in the values in it:
[tex]Ka=\frac{(0.003)^2}{(0.1-0.003)}[/tex]
0.003 is almost negligible as compared to 0.10, so (0.10 - 0.003) could be taken as 0.10.
[tex]Ka=\frac{(0.003)^2}{(0.01)}[/tex]
[tex]Ka=9*10^-^5[/tex]
Second choice is the right one.
