Question

Calculate the [H3O+] of 0.35 M solution of benzoic acid HC7H3O2

Answer 1

Answer:

$[H_3O^+]=4.7x10^{-3}M$

Explanation:

Hello there!

In this case, since the ionization of benzoic acid is:

$HC_7H_3O_2+H_2O\rightleftharpoons C_7H_3O_2^-+H_3O^+$

Whereas the corresponding equilibrium expression is:

$Ka=\frac{[C_7H_3O_2^-][H_3O^+]}{[HC_7H_3O_2]}$

Now, we insert the acidic equilibrium constant and the reaction extent x, to write:

$6.3x10^{-5}=\frac{x^2}{0.35M}$

Thus, by solving for x, we obtain:

$x=\sqrt{6.3x10^{-5}*0.35}\\\\x=4.7x10^{-3}M$

Which is also:

$x=4.7x10^{-3}M$

Best regards!