Question

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

Answer 1

`[H_3O^+]=4.7x10^(-3)M`

Step-by-step 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=([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)=(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!