Guest prvbspike Posted February 23, 2004 Share Posted February 23, 2004 alright.....this is my chem hw tonight, and i dont really know if i am doing it right....so i have sodium benzoate, C6H5COONa, a 0.10 M solution of it, and it has a pH of 8.60 at room temp, i need to calculate the Ka for benzoic acid, C6H5COOH...the [OH-] is 4x10^-6, and the equilibrium constant for the rxn: C6H5COO- + H2O--> C6H5COOH + OH- is 1.6x10^-10......i want to find the Ka of benzoic acid....HELP PLEASE! Link to comment Share on other sites More sharing options...
Cookie Posted February 23, 2004 Share Posted February 23, 2004 I think you need to use the Henderson Hasselbach equation. Been awhile since I've done acid-base chemistry though. Cookie Link to comment Share on other sites More sharing options...
agaubr Posted February 27, 2004 Share Posted February 27, 2004 Ka X Kb = Kw so if you know one or the other can caluate the since Kw = 1 X 10-14. Link to comment Share on other sites More sharing options...
wolfson Posted February 27, 2004 Share Posted February 27, 2004 Thus: Ka = 6.5e-5 Link to comment Share on other sites More sharing options...
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