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How do I prepare 1M NaOH 1L (assay 98%)?
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+ Sodium hydroxide
+ Chemistry
+ Organic chemistry
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Linda C. Ragin
How do I prepare 1M NaOH 1L (assay 98%)?
First of all prepare 0.1 normal or molar NaOH solution by dissolving 4 gm NaOH in 1000 ml or 2 gm NaOH in 500 ml of distilled water. Now dissolve 2 gm of Na2CO3 in 100 ml 0.1 normal or molar NaOH solution prepared above.
First of all prepare 0.1 normal or molar NaOH solution by dissolving 4 gm NaOH in 1000 ml or 2 gm NaOH in 500 ml of distilled water. Now dissolve 2 gm of Na2CO3 in 100 ml 0.1 normal or molar NaOH solution prepared above.
Because assay is 98% , mass required = 39.9971*100/98 = 40.81337 g
IN THEORY:
Weigh out 40.81337g of the NaOH and dissolve in some distilled water . Transfer quantitatively to a 1.0L volumetric flask and make up to the mark with additional distilled water . You have a 1.0L of 1.0M NaOH solution.
WHY : did I write IN THEORY?
Because in practice you cannot do this . NaOH is hygroscopic - it absorbs water from the water vapour in the air . It also reacts with the CO2 in the air. During the weighing out process the NaOH is constantly changing mass as it absorbs H2O and CO2 .
Therefore you will find it impossible to weigh out any exact mass of NaOH.
The best you can do is: Rapidly weigh out , as clasely as possible +/- 40.8g of the NaOH - any mass between 40.5 and 41.0g will be OK
Proceed to dissolve this NaOH and make up to 1.0L in a volumetric flask.
Then you have to find the exact molarity of your solution by standardising it against a primary standard such as KHP.
It is common in laboratories to see reagent bottles of standard NaOH solutions marked:
1.0M NaOH solution ( f = 0.995) In this case the “1.0M NaOH” solution has a corrected molarity of 0.995M
Because assay is 98% , mass required = 39.9971*100/98 = 40.81337 g
IN THEORY:
Weigh out 40.81337g of the NaOH and dissolve in some distilled water . Transfer quantitatively to a 1.0L volumetric flask and make up to the mark with additional distilled water . You have a 1.0L of 1.0M NaOH solution.
WHY : did I write IN THEORY?
Because in practice you cannot do this . NaOH is hygroscopic - it absorbs water from the water vapour in the air . It also reacts with the CO2 in the air. During the weighing out process the NaOH is constantly changing mass as it absorbs H2O and CO2 .
Therefore you will find it impossible to weigh out any exact mass of NaOH.
The best you can do is: Rapidly weigh out , as clasely as possible +/- 40.8g of the NaOH - any mass between 40.5 and 41.0g will be OK
Proceed to dissolve this NaOH and make up to 1.0L in a volumetric flask.
Then you have to find the exact molarity of your solution by standardising it against a primary standard such as KHP.
It is common in laboratories to see reagent bottles of standard NaOH solutions marked:
1.0M NaOH solution ( f = 0.995) In this case the “1.0M NaOH” solution has a corrected molarity of 0.995M
First of all prepare 0.1 normal or molar NaOH solution by dissolving 4 gm NaOH in 1000 ml or 2 gm NaOH in 500 ml of distilled water. Now dissolve 2 gm of Na2CO3 in 100 ml 0.1 normal or molar NaOH solution prepared above.
First of all prepare 0.1 normal or molar NaOH solution by dissolving 4 gm NaOH in 1000 ml or 2 gm NaOH in 500 ml of distilled water. Now dissolve 2 gm of Na2CO3 in 100 ml 0.1 normal or molar NaOH solution prepared above.
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Molar mass of NaOH is 39.9971g/mol
Because assay is 98% , mass required = 39.9971*100/98 = 40.81337 g
IN THEORY:
Weigh out 40.81337g of the NaOH and dissolve in some distilled water . Transfer quantitatively to a 1.0L volumetric flask and make up to the mark with additional distilled water . You have a 1.0L of 1.0M NaOH solution.
WHY : did I write IN THEORY?
Because in practice you cannot do this . NaOH is hygroscopic - it absorbs water from the water vapour in the air . It also reacts with the CO2 in the air. During the weighing out process the NaOH is constantly changing mass as it absorbs H2O and CO2 .
Therefore you will find it impossible to weigh out any exact mass of NaOH.
The best you can do is: Rapidly weigh out , as clasely as possible +/- 40.8g of the NaOH - any mass between 40.5 and 41.0g will be OK
Proceed to dissolve this NaOH and make up to 1.0L in a volumetric flask.
Then you have to find the exact molarity of your solution by standardising it against a primary standard such as KHP.
It is common in laboratories to see reagent bottles of standard NaOH solutions marked:
1.0M NaOH solution ( f = 0.995) In this case the “1.0M NaOH” solution has a corrected molarity of 0.995M
Molar mass of NaOH is 39.9971g/mol
Because assay is 98% , mass required = 39.9971*100/98 = 40.81337 g
IN THEORY:
Weigh out 40.81337g of the NaOH and dissolve in some distilled water . Transfer quantitatively to a 1.0L volumetric flask and make up to the mark with additional distilled water . You have a 1.0L of 1.0M NaOH solution.
WHY : did I write IN THEORY?
Because in practice you cannot do this . NaOH is hygroscopic - it absorbs water from the water vapour in the air . It also reacts with the CO2 in the air. During the weighing out process the NaOH is constantly changing mass as it absorbs H2O and CO2 .
Therefore you will find it impossible to weigh out any exact mass of NaOH.
The best you can do is: Rapidly weigh out , as clasely as possible +/- 40.8g of the NaOH - any mass between 40.5 and 41.0g will be OK
Proceed to dissolve this NaOH and make up to 1.0L in a volumetric flask.
Then you have to find the exact molarity of your solution by standardising it against a primary standard such as KHP.
It is common in laboratories to see reagent bottles of standard NaOH solutions marked:
1.0M NaOH solution ( f = 0.995) In this case the “1.0M NaOH” solution has a corrected molarity of 0.995M
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