... volume of NaOH added. Theoretically, this should give me a pH curve from which to work. In order to obtain a value for Ka, I shall find the equivalence point (at around 25cm3 where the graph goes up vertically) and find the pH at this point, I shall then divide this by two to give me the 1/2 equivalence point. Having obtained this value, I can assume that: [CH3COOH] = [CH3COO-] = [H3+O] And as: Ka = [CH3COO-] [H3+O] [CH3COOH] Therefore: Ka = [H3+O] Or: pKa = pH (at 1/2 equivalence point) Or: pH = -Log10 Ka Prediction I know that the curve for a weak acid/strong alkali should start higher up the axis than a strong acid/strong base curve. This results in the vertical section of the graph being shorter and due to salt hydrolysis, the curve will rise to 14 far quicker than it should. This is due to the following mechanism: CH3COOH + NaOH › CH3COONa + H2O CH3COONa › CH3COO- ...

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