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Author Topic: Molality question  (Read 789 times)
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needy
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« on: December 13, 2009, 03:42:42 AM »
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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Borek
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« Reply #1 on: December 13, 2009, 05:47:04 AM »
Quote from: needy on December 13, 2009, 03:42:42 AM
change KHP to moles = 0.00403 g KHP = 0.00403 g NaOH

g? Or moles?

Quote
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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needy
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« Reply #2 on: December 13, 2009, 06:08:56 AM »
oops you are right it should have been 0.00403 mol NaOH.  But what is the mass of solvent?
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Borek
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« Reply #3 on: December 13, 2009, 08:01:05 AM »
Solution consist of solvent and solute.
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« Reply #4 on: December 13, 2009, 09:35:20 AM »
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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« Reply #5 on: December 13, 2009, 10:15:18 AM »
That's right
Proceed with the next part. Smiley
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:-) SKS
needy
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« Reply #6 on: December 13, 2009, 10:28:50 AM »
You guys are the best, off to email my class!!
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