you are asked to prepare a ph=4.00 buffer starting from 1.50 L of 0.200 M soluti

Question

you are asked to prepare a ph=4.00 buffer starting from 1.50 L of 0.200 M solution of benzoic acid ( C6H5COOH) and an excess of sodium benzoate ( C6H5COONa) A). what is the pH of the benzoic acid solution prior to adding sodium benzoate ?? B). how many grams of sodium benzoate should be added to prepare the buffer ? Neglect the small volume change the occurs when the sodium benzoate is added

Explanation / Answer

pKa(CH3COOH) = 4.202 ................=> Ka=6.28 x 10^-5 .......... X^2 / (0.2-X) =6.28 x 10^-5 ...............=> X = 0.00354 M .......=>a) pH = - log 0.00354 = 2.45 b) pH = pKa + log (S/a) ........=> 4 = 4.202 + log ( S / 0.2) .........=> S = 0.1256 M ............=> Moles = 0.1256 x 1.5 = 0.1884 moles => Grams = 0.1884 x 82 = 15.45 grams

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