Answer:
[H⁺] = 4.56x10⁻³ M
pH = 2.34
Explanation:
Hello there!
In this case, according to the ionization reaction of benzoic acid:
[tex]HC_6H_5CO_2+H_2O\rightleftharpoons C_6H_5CO_2^++H_3O^+[/tex]
Whereas [tex][H_3O^+]=[H^+][/tex], we can set up the equilibrium expression in terms of [tex]x[/tex] (reaction extent) to obtain:
[tex]Ka=\frac{[C_6H_5CO_2^-][H_3O^+]}{[HC_6H_5CO_2]} \\\\6.5x10^{-5}=\frac{x^2}{0.32-x}[/tex]
However, since Ka<<<1, we can neglect the [tex]x[/tex] on bottom to easily solve for it:
[tex]6.5x10^{-5}=\frac{x^2}{0.32}\\\\x=\sqrt{6.5x10^{-5}*0.32} \\\\x=4.56x10^{-3}[/tex]
Which is actually the same as [H⁺]. Finally, the pH turns out to be:
[tex]pH=-log(4.56x10^{-3})\\\\pH=2.34[/tex]
Best regards!
