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### Topic: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH  (Read 20610 times)

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#### dankamus

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##### How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« on: March 16, 2016, 10:04:51 PM »
Me and a coworker were arguing about this today. He says that you'd need to use M1V1 = M2V2 --> assuming 14.8 N ~14.8 M then 14.8 M NH4OH(V2) = 0.1 M NH4OH(500 mL) = 3.38 mL of 29.7% NH4OH. I found an online calculator that says you'd actually need 6.57 mL of NH4OH. My own calculation gave the same result and other sites I've found support this # too.

First off which is right?

Second, based on my understanding, the only way that 29.7% NH4OH could be 14.8 N is if the 14.8 is reported as NH3. Why would a label for 29.7% NH4OH report the normality as NH3, when the molecular formula is reported as NH4OH? Is this just convention, or am I missing something?

#### Hunter2

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #1 on: March 17, 2016, 02:33:26 AM »
There is nothing wrong in the calculation. The 3.38 ml is calculated for 500 ml and the 6.57 for 1 liter solution.

The Mole of NH3 is equal to the mole of NH4OH if we assume  the reaction goes 100% NH3 + H2O => NH4OH what is not the case .

Normal the percantager should be giveb as dissolved NH3.

#### Dan

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #2 on: March 17, 2016, 03:11:18 AM »
Second, based on my understanding, the only way that 29.7% NH4OH could be 14.8 N is if the 14.8 is reported as NH3. Why would a label for 29.7% NH4OH report the normality as NH3, when the molecular formula is reported as NH4OH? Is this just convention, or am I missing something?

I too was confused by this before. In the lab, we had a bottle of "35% ammonium hydroxide" and there was some confusion as to whether this was 35% based on the mass of NH4OH or NH3 (the molarities of which would differ by a factor of ~2). Long story short, titrated showed it was 35% ammonia basis.

3.38 mL is the correct answer for your case. Can you show your calculations (and the online calculators) that arrived at 6.57 mL?
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#### Borek

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #3 on: March 17, 2016, 03:54:54 AM »
29.7% w/w is more like 15.6 M, so the volume needed is a bit lower. I got 3.20 mL.

For most practical applications the difference doesn't matter.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### dankamus

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #4 on: March 17, 2016, 08:03:47 AM »
There is nothing wrong in the calculation. The 3.38 ml is calculated for 500 ml and the 6.57 for 1 liter solution.

The Mole of NH3 is equal to the mole of NH4OH if we assume  the reaction goes 100% NH3 + H2O => NH4OH what is not the case .

Normal the percantager should be giveb as dissolved NH3.

I was actually using 500mL for both calculations and 3.38*2 doesn't equal 6.57, it's 6.76. The numbers are similar but are based on the molecular weight of either NH3 or NH4OH.

Second, based on my understanding, the only way that 29.7% NH4OH could be 14.8 N is if the 14.8 is reported as NH3. Why would a label for 29.7% NH4OH report the normality as NH3, when the molecular formula is reported as NH4OH? Is this just convention, or am I missing something?

I too was confused by this before. In the lab, we had a bottle of "35% ammonium hydroxide" and there was some confusion as to whether this was 35% based on the mass of NH4OH or NH3 (the molarities of which would differ by a factor of ~2). Long story short, titrated showed it was 35% ammonia basis.

3.38 mL is the correct answer for your case. Can you show your calculations (and the online calculators) that arrived at 6.57 mL?

This is actually what we'd planned on doing. I still might just out of curiosity.

My calc: 29.7% w/w * 0.9 g/mL = 261 g/L as NH4OH
--> 261 g/L / 34 g/mol = 7.68 mol NH4OH
--> 7.68 M NH4OH (x) = 0.1 M NH4OH (500 mL)
= 6.51 mL 7.68 M NH4OH

Sigma online calculator: http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html

Another source I found that just shows how much to use w/no calcs: chemistry.about.com/od/labrecipes/a/baserecipes.htm

29.7% w/w is more like 15.6 M, so the volume needed is a bit lower. I got 3.20 mL.

For most practical applications the difference doesn't matter.

The bottle actually says 29.7% w/w and 14.8 N.

Thanks.

#### Borek

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #5 on: March 17, 2016, 10:49:05 AM »
29.7% w/w is more like 15.6 M, so the volume needed is a bit lower. I got 3.20 mL.

For most practical applications the difference doesn't matter.

The bottle actually says 29.7% w/w and 14.8 N.

Density tables say that solution that is 29.7% w/w has a density of 0.896 g/mL. That means 1 L weighs 896 g, of that 0.297*896=266 g is ammonia. That's 15.6 moles.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### dankamus

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #6 on: March 17, 2016, 12:15:57 PM »
So I finally solved this problem. The 29.7% listed on the bottle of NH4OH we have is 29.7% as NH3, not NH4OH. Apparently it's convention to list it this way? Anyway both the % w/w and the normality listed are NH3, even though the bottle says NH4OH.

I realized that if you deliver more than 3.38 mL you're actually giving an overdose of NH4OH.

#### Dan

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##### Re: How to make 0.1 M NH4OH from 29.7% (14.8 N) NH4OH
« Reply #7 on: March 19, 2016, 07:25:08 AM »
So I finally solved this problem. The 29.7% listed on the bottle of NH4OH we have is 29.7% as NH3, not NH4OH. Apparently it's convention to list it this way? Anyway both the % w/w and the normality listed are NH3, even though the bottle says NH4OH.

Yes, exactly as I described in Reply #2 above.

I don't know why it's listed that way, seems illogical to me.
My research: Google Scholar and Researchgate