Posted by tomi on Thursday, April 2, 2009 at 12:13pm.
The pH of the satd soln is 13.12; therefore, the pOH = 14 - 13.12 = 0.88
From this and pOH = -log(OH^-) you can find (OH^-) = about 0.12 (you need to do it exactly).
This was diluted from 15 mL to 250 so the (OH^-) in the 250 mL flask is 0.12 x 15/250 = about 0.0072 M (again you need to go through it to get a more exact answer).
So now you have a 15.0 mL sample of 0.0072 M OH^- titrated with 29.3 mL of ?M HCl.
mLbase x M base = mLacid x M acid.
Solve for M acid.
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