# Chemistry

Find the pH experimentaly .1 ml before and .1 ml after the equivalence point if the experimental equivalence point was found to be at 35 ml. In this titration, you are titrating 0.500 M NaOH with 0.250 M HCl.

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1. A complicated work but do-able.

NaOH + HCl ==> NaCl + H2O.

I think one must decide where we are at the equivalence point.
Since the M NaOH is twice that of the HCl, that means the HCl will take twice as much to neutralize the NaOH.
So that 35 mL is made up of 2 mL HCl for every mL NaOH. The way I see it is that the equivalence point occurs at 11.67 mL base and 23.34 mL HCl.
That gives us what in the way of millimoles?
11.67 x 0.500 = 5.835 mmoles NaOH.
23.34 x 0.26 = 5.835 mmoles HCl.

Now calculate mmoles NaOH untitrated at 23.24 mL HCl, convert to M, then pOH, then pH.

Then calculate mmols HCl excess at 23.44, convert to M, then to pH.

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posted by DrBob222

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