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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Cavalier on December 03, 2010, 11:12:09 AM

Title: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: Cavalier on December 03, 2010, 11:12:09 AM
Ok...  I took Chem classes, but that was quite a while ago.  I need some help understanding this statement....

"20% aqueous sodium hydroxide (NaOH)"

If I have dry crystal NaOH that I purchased from Home Depot, how do I turn that into 20% aqueous sodium hydroxide.  And in particular, how do I do this without dissolving my garage or kitchen.  Does the % mean by weight?  20 grams of NaOH for every 100 grams of water?

Thank you all in advance.

RC
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: Borek on December 03, 2010, 05:21:26 PM
20g of NaOH per 100g of SOLUTION, not water.
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: bberti on January 03, 2011, 07:05:00 PM
Assuming your 20% is weight on weight, just add 80 grams of water (80 mL at 20 degrees Celsius) to 20 grams of NaOH.
You'll have [(20g NaOH) /(20g NaOH+80g water) ]*100 ==> 20g NaOH every 100g of solution.
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: darko on January 04, 2011, 05:53:48 AM
Quote from: bberti on January 03, 2011, 07:05:00 PM
Assuming your 20% is weight on weight, just add 80 grams of water (80 mL at 20 degrees Celsius) to 20 grams of NaOH.
You'll have [(20g NaOH) /(20g NaOH+80g water) ]*100 ==> 20g NaOH every 100g of solution.

if you atempt to add temperature in calculation (in this case is better not to), then is more precise 4oC, ;D

i think that is more imprtant to explain that it is egzothermic reaction
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: vmelkon on January 23, 2011, 05:36:22 PM
Quote from: darko on January 04, 2011, 05:53:48 AM
Quote from: bberti on January 03, 2011, 07:05:00 PM
Assuming your 20% is weight on weight, just add 80 grams of water (80 mL at 20 degrees Celsius) to 20 grams of NaOH.
You'll have [(20g NaOH) /(20g NaOH+80g water) ]*100 ==> 20g NaOH every 100g of solution.

if you atempt to add temperature in calculation (in this case is better not to), then is more precise 4oC, ;D

i think that is more imprtant to explain that it is egzothermic reaction

egzothermic or egg-zothermic. Nah, exothermic.
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: darko on January 24, 2011, 10:41:23 AM
Quote from: vmelkon on January 23, 2011, 05:36:22 PM
Quote from: darko on January 04, 2011, 05:53:48 AM
Quote from: bberti on January 03, 2011, 07:05:00 PM
Assuming your 20% is weight on weight, just add 80 grams of water (80 mL at 20 degrees Celsius) to 20 grams of NaOH.
You'll have [(20g NaOH) /(20g NaOH+80g water) ]*100 ==> 20g NaOH every 100g of solution.

if you atempt to add temperature in calculation (in this case is better not to), then is more precise 4oC, ;D

i think that is more imprtant to explain that it is egzothermic reaction

egzothermic or egg-zothermic. Nah, exothermic.



 :-*
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: Lord Ferrum on January 26, 2011, 01:40:13 AM
That's why it's easier to use moles per liter  ;)
Title: Re: 20% aqueous sodium hydroxide (NaOH) - Need Help Understanding.
Post by: skbuncks on January 26, 2011, 06:36:39 AM
Quote from: Lord Ferrum on January 26, 2011, 01:40:13 AM
That's why it's easier to use moles per liter  ;)

Not if your doing it in your garage without access to a volumetric flask it isn't.

skb