Homework Help: chemistry

Posted by Dustin on Monday, April 7, 2008 at 5:15pm.

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, Monday, April 7, 2008 at 6:22pm
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, Monday, April 7, 2008 at 6:49pm
0.003898-- this is H+?

Ka = ( )(0.00398)/( )

What would be the concentration of C6H5COO- ?
chemistry - DrBob222, Monday, April 7, 2008 at 7:04pm
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, Tuesday, April 8, 2008 at 1:40am
I get 4.0 x 10^-3 but im incorrect
chemistry - DrBob222, Thursday, April 10, 2008 at 1:29pm
No.
(0.00398)^2/(0.25-0.00398) =
6.44 x 10^-5

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