Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: needy on December 13, 2009, 09:42:42 AM
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It has been awhile since I did a molality question, if someone could check my Math I would greatly appreciate it.
Titration of 0.824 grams of KHP (204.22g/mol) required 38.314 grams of NaOH solution to reach a phenolphthalein endpoint. Find the molality of the NaOH solution .
This is what I did. change KHP to moles = 0.00403 g KHP = 0.00403 g NaOH
Molality = 0.00403 g NaOH/0.03814 Kg = 0.106 m
Based upon the solution find the concentration of H2SO4 solution in mol/kg if a 10.063 g aliquot of H2SO4 solution required 57.911 grams of NaOH solution to reach the phenolphthalein endpoint.
m x 0.010036 kg = 0.106 moles x 1 kg
moles of H2SO4 = 10.5 moles
m=10.5 moles H2SO4/0.010063 kg + 1 kg = 10.4 m H2SO4
If I did it wrong any hints would be appreciated.
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change KHP to moles = 0.00403 g KHP = 0.00403 g NaOH
g? Or moles?
Molality = 0.00403 g NaOH/0.03814 Kg = 0.106 m
No, 0.038314 is a mass of solution, not solvent.
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oops you are right it should have been 0.00403 mol NaOH. But what is the mass of solvent?
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Solution consist of solvent and solute.
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Oh boy I am so out of it, what was I thinking going back to school at my age? Ok last attempt.
0.0043 moles NaOH = 0.161 g NaOH
38.314 grams of solution NaOH - 0.161 g solute NaOH = 38.153 g solvent
so maybe then this is correct
m=0.00403 moles of NaOH/0.038153 kg = 0.106 m NaOH
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That's right
Proceed with the next part. :)
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You guys are the best, off to email my class!!