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Topic: molar mass problem  (Read 7824 times)

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ahsgpc07

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molar mass problem
« on: August 15, 2005, 09:34:20 PM »
The question reads: A 2.29 g sample of an unknown acid is dissolved in 1.0 liter of water. A titration required 25.0 ml of 0.500 M NaOH to completely react with all the acid present. What is the molar mass of the acid?

I'm not really looking for an answer,  but i cant figure out the where to start. I have started by finding the moles of NaOH required (.0125) and now i'm pretty lost..

Offline Qazzian

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Re:molar mass problem
« Reply #1 on: August 15, 2005, 10:40:50 PM »
Well, you're on the right track.

You can't find the molarity of the acid, but you can find the normality. The difference between the two is basically the molarity is moles/L, and the normality is charge/L. For acids like HCl and nitric acid which only have one H+, these two are the same, but H2SO4 is different.

Ok, back on track. You have the number of moles of NaOH that are used. Now, assume a 1:1 ratio. That means you need the same number of moles of acid as you used of base. You have the number of moles of the acid, and the number of grams was given, so you can find the molar masds by dividing one by the other.

When you come across problems like this, it's always helpful to look at the units you want (in this case g/mol), and look at what you're given in the question. That way if you get stuck, you may have something to help you out.
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Offline Borek

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Offline jubba

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Re:molar mass problem
« Reply #3 on: August 16, 2005, 07:12:42 AM »
wouldn't the answer be different depending on whether the acid is monoprotic diprotic ... etc

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Re:molar mass problem
« Reply #4 on: August 16, 2005, 08:08:49 AM »
wouldn't the answer be different depending on whether the acid is monoprotic diprotic ... etc

Will be. That's why you can determine only mass of equivalent, not molar mass. For molar mass you need additional information - for example from freezing point depression measurement.
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camur

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Re:molar mass problem
« Reply #5 on: August 16, 2005, 10:58:19 AM »
If we think that the acid is monoprotic

HX+NaOH---> NaX + H2O

the number of moles of NaOH is  0.5*0.025=0.0125 moles

the number of moles of the acid is also equal to 0.0125

2.29/0.0125 =183.2 g/mole(if the acid is monoprotic)

Offline Qazzian

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Re:molar mass problem
« Reply #6 on: August 16, 2005, 11:18:36 AM »
thanks Borek. My mind has turned to Jello, couldn't remember the right word for it.
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Blueshawk

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Re:molar mass problem
« Reply #7 on: August 19, 2005, 02:40:47 PM »
Camur,

with a molar mass of  183 g/mol    don't you think that is a little high for most monoprotic acids in high school chemistry.?

To me its seems more likely to be a diprotic acid, which usually have a higher molar mass, but not usually above 150 g/mol ???

H2SO4 = 98 g/mol

Unless it is a hydrocarbon acid.... some sort of carboxylic acid ???  just an after thought.

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