Question

Benzoic acid, HC6H5CO2,is a monoprotic acid(only one H+ ionizes)with a Ka=6.5×10^-5. Calculate [H+] and the pH of a .32M solution of benzoic acid. PLEASE ANSWER. ​

Answer 1

[H⁺] = 4.56x10⁻³ M

pH = 2.34

Step-by-step explanation:

Hello there!

In this case, according to the ionization reaction of benzoic acid:

`HC_6H_5CO_2+H_2O\rightleftharpoons C_6H_5CO_2^++H_3O^+`

Whereas
`[H_3O^+]=[H^+]`, we can set up the equilibrium expression in terms of
`x` (reaction extent) to obtain:

`Ka=([C_6H_5CO_2^-][H_3O^+])/([HC_6H_5CO_2]) \\\\6.5x10^(-5)=(x^2)/(0.32-x)`

However, since Ka<<<1, we can neglect the
`x` on bottom to easily solve for it:

`6.5x10^(-5)=(x^2)/(0.32)\\\\x=\sqrt{6.5x10^(-5)*0.32} \\\\x=4.56x10^(-3)`

Which is actually the same as [H⁺]. Finally, the pH turns out to be:

`pH=-log(4.56x10^(-3))\\\\pH=2.34`

Best regards!