Posted by Dustin on .
A 0.25 mol/L solution of benzoic acid, KC7H5O2 and antiseptic also used as a food preservative, has a pH of 2.40. Calculate the Ka of benzoic acid at SATP
My work:
C6H5COOH > C6H5COO^ + H^+
Ka = (C6H5COO)(H+)/ C6H5COOH
I know my next steps are to use pH = log(H+) to calculate H+
and then
substitute into Ka expression and determine Ka
How to do this?
Thanks in advance

chemistry 
DrBob222,
pH = 2.40
pH = log(H+)
log(H^+) = 2.40
log(H^+) = 2.40
plug 2.40 into your calculator and hit the 10^{x} key to obtain 0.00398. check my work. 
chemistry 
Ross,
0.003898 this is H+?
Ka = ( )(0.00398)/( )
What would be the concentration of C6H5COO ? 
chemistry 
DrBob222,
Look at the equation.
C6H5COOH ==> C6H5COO^ + H^+
For every 1 mol C6H5COOH that ionizes, you get 1 mol C6H5COO^ and 1 mol H^+. So they are the same. 
chemistry 
Kelsy,
I get 4.0 x 10^3 but im incorrect

chemistry 
DrBob222,
No.
(0.00398)^2/(0.250.00398) =
6.44 x 10^5