Monday
March 30, 2015

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

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