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Answer:
[tex][H_3O^+]=4.7x10^{-3}M[/tex]
Explanation:
Hello there!
In this case, since the ionization of benzoic acid is:
[tex]HC_7H_3O_2+H_2O\rightleftharpoons C_7H_3O_2^-+H_3O^+[/tex]
Whereas the corresponding equilibrium expression is:
[tex]Ka=\frac{[C_7H_3O_2^-][H_3O^+]}{[HC_7H_3O_2]}[/tex]
Now, we insert the acidic equilibrium constant and the reaction extent x, to write:
[tex]6.3x10^{-5}=\frac{x^2}{0.35M}[/tex]
Thus, by solving for x, we obtain:
[tex]x=\sqrt{6.3x10^{-5}*0.35}\\\\x=4.7x10^{-3}M[/tex]
Which is also:
[tex]x=4.7x10^{-3}M[/tex]
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