Answer:
\(Ka=1.53x10^-5\)
Explanation:
Since it is a weak acid it is necessary to write the balance. Where the unknown acid will be HA.
\(HA<->\hspace{0.1cm}A^- \hspace{0.1cm}+\hspace{0.1cm}H^+\).
Then with the equilibrium we can write the ICE table (figure 1).
Where we obtain the equation:
\(Ka=\frac{X*X}{0.3-X} =\frac{X^2}{0.3-X}\)
Now, we have to find "X". To do this we can use the pH equation and the pH value:
\(pH=-Log\hspace{0.1cm}H^+\)
\(-pH=Log\hspace{0.1cm}H^+\)
\(10^-pH=H^+\)
\(10^-2.67=H^+\)
\(H^+=0.00213\)
\(X=0.00213\)
Then we have to plug the values in the Ka equation:
\(Ka=\frac{(0.00213)^2}{0.3-0.00213}\)
\(Ka=1.53x10^-5\)
