Posted by marla on Monday, February 6, 2012 at 3:31pm.
There are two ways, one is an estimate and the other is exact.
For the estimate, look at the pH values. When the pH changes from low numbers (you said you were titrating HCl by adding 1 mL portions of NaOH) to high numbers (pH 2 or 3 to pH 9 or 10), the equivalence point is between those two values of NaOH. Then mL x M = mL x M and solve for M NaOH.
The exact way is not nearly that easy. Take the second derivative of the additions, plot the second derivative of pH versus mL. The pH will curve will change sharply from a high number to aa low number. Where the line crosses the zero point is the mL for the equivalence point. See if that method is described in your text.
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