Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: shakajjh on February 10, 2008, 02:25:23 AM
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Ok so I did this problem about 5 times with 3x different data and what the professor showed us in class and the answer in the back of the solutions manual does not match up
The following question refers to the titration of 25.00 mL of 0.132 M HCl with 0.128 M Ba(OH) 2. The reaction proceeds according to the following stoichiometry:
2HCl + Ba(OH)2 ---- > BaCl2 + 2H2O
What will the pH of the titrated solution be after the addition of 5.00 mL of Ba(OH)2? NOTE: use two significant figures in your answer.
IN addition to this question how can I find the pH of the titrated solution at the equivalence point?
So here is solution according to professor
we first do mmol of HCl = 25.00 x 0.132 = 3.30 mmol HCl
& mmol Ba(OH)2 = 5.00 ML x .128 M Ba(OH)2 ok here my professor multipled by stoic ratio of 2 HCl/1mmol(OH)2 to get 1.28 mmol of HCl.
He then took the difference between the 3.30 mmol HCL and the 1.28 mmol of HCl which got it to be 2.02 mmol of HCl.
then [HCl] = 2.02 mmol/30.00 ml = 0.0673333333
then PH = -log [.067333333]
which is equal to 1.17
in the solutions manual they don't seem to take into consideration the stoic ratio for this and that gave the correct answer in there.
I have to submit this problem exactly online so therefore Im confused about the right answer
thanks for all your help :)
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in the solutions manual they don't seem to take into consideration the stoic ratio for this and that gave the correct answer in there.
1.17 is the correct answer, you have to take stoichiometry into account when dealing with multiprotic acids/bases.