posted by Dustin .
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
C6H5COOH --> C6H5COO^- + H^+
Ka = (C6H5COO-)(H+)/ C6H5COOH
I know my next steps are to use pH = log(H+) to calculate H+
substitute into Ka expression and determine Ka
How to do this?
Thanks in advance
pH = 2.40
pH = -log(H+)
-log(H^+) = 2.40
log(H^+) = -2.40
plug -2.40 into your calculator and hit the 10x key to obtain 0.00398. check my work.
0.003898-- this is H+?
Ka = ( )(0.00398)/( )
What would be the concentration of C6H5COO- ?
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.
I get 4.0 x 10^-3 but im incorrect
6.44 x 10^-5