Respuesta :
Answer : The equilibrium concentration of [tex]H_3O^+[/tex] in the solution is, [tex]2.1\times 10^{-3}M[/tex]
Explanation :
The dissociation of acid reaction is:
[tex]C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-[/tex]
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = [tex]7.0\times 10^{-2}M[/tex]
[tex]K_a=6.3\times 10^{-5}[/tex]
The expression of dissociation constant of acid is:
[tex]K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}[/tex]
[tex]K_a=\frac{(x)\times (x)}{(c-x)}[/tex]
Now put all the given values in this expression, we get:
[tex]6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}[/tex]
[tex]x=2.1\times 10^{-3}M[/tex]
Thus, the equilibrium concentration of [tex]H_3O^+[/tex] in the solution is, [tex]2.1\times 10^{-3}M[/tex]
