Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: pragathi on August 06, 2008, 07:21:36 AM
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how to prepare 2%nitric acid from 67-70%nitric acid?
How many ml we need to take from 67-70%nitric acid?
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68.5/2 = ?
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2% v/v = 2mL pure HNO3 per 100 mL water
1000 mL of 2% v/v HNO3 contains 20 mL of 100% HNO3
therefore if the HNO3 is 67% pure it will require 20/(67/100)=29.9 mL
if the HNO3 is 70% pure it will require 20/(70/100)=28.6 mL
Mean value = 29.25 mL
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2% v/v
Why v/v? Question didn't ask for v/v, most likely it was about w/w. Not that it changes the result much.
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2% v/v
Why v/v? Question didn't ask for v/v, most likely it was about w/w. Not that it changes the result much.
Just an assumption as the HNO3 is supplied as a solution and the % content on spec sheet from supplier is based on liquid titration
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You guys sure are making it complicated for a solution that is 67-60% pure. ;)
The average value, 68.5/2 (you want to go from 68.5% to 2%) gives a dilution factor of 34.25.
1000mL/34.25 = 29.2 mL
Which is essentially the same answer JGK's method get when you round after the math.
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What if I tell you that you need 21 mL per L?
What is the density of stock nitric acid?
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Yeah, but you changes to w/w. Given the vagueness of the question I went with v/v, as did JGK!