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.
Quote from:
Quote from:
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
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.
Quote from:
Quote from:
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
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.
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.
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.
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.
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.
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.
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?
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?
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.
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.
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VOTE
For most practical applications the difference doesn't matter.
For most practical applications the difference doesn't matter.
More
VOTE
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.
Quote from:
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
Quote from:
The bottle actually says 29.7% w/w and 14.8 N.
Thanks.
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.
Quote from:
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
Quote from:
The bottle actually says 29.7% w/w and 14.8 N.
Thanks.
More
VOTE
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.
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.
More
VOTE
Yes, exactly as I described in Reply #2 above.
I don't know why it's listed that way, seems illogical to me.
Yes, exactly as I described in Reply #2 above.
I don't know why it's listed that way, seems illogical to me.
More
VOTE
I realized that if you deliver more than 3.38 mL you're actually giving an overdose of NH4OH.
I realized that if you deliver more than 3.38 mL you're actually giving an overdose of NH4OH.
More
VOTE
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?
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?
More
VOTE