i used to do experiments during my master. I used the weight measurement to make the solution. Let's say for 1l solution i.e. 1000ml. If you Want to make a 10% solution. Add 100 g of NaOH to 900ml (=900g) of water (De ionized). That will give you a 10% NaOH solution.
i used to do experiments during my master. I used the weight measurement to make the solution. Let's say for 1l solution i.e. 1000ml. If you Want to make a 10% solution. Add 100 g of NaOH to 900ml (=900g) of water (De ionized). That will give you a 10% NaOH solution.
There is no "chemical reaction" but there ia a reaction in that the disassociation of the sodium and hydroxide ions and the hydration of those ions releases a LOT of heat, enough to boil water in some circumstances. Great care is needed.
There is no "chemical reaction" but there ia a reaction in that the disassociation of the sodium and hydroxide ions and the hydration of those ions releases a LOT of heat, enough to boil water in some circumstances. Great care is needed.
Yes but everyone is forgetting that every reaction is an equilibrium. This would mean that it would depend on the concentration of sodium hydroxide solution. The pKa of ethanol is 15.9 which if you assume a 1M NaOH solution the pH would be about 14 (not including minute amounts of undisaccociated base). PH=pKa+log(A-/HA) -1.9=log(A-/HA) 0.0126 = (CH3CH2O-/CH3CH2OH). This means that the percent of ethoxide ions that are in solution is just over 1% (I can go into % ionization if you would like) which is certainly detectable. Similarly a 10M (difficult to achieve) solution of sodium hydroxide (pH=15) would have just over 10% ethoxide ions vs ethanol. The reaction occurs, but only under a weak equilibrium depending on concentration.
Yes but everyone is forgetting that every reaction is an equilibrium. This would mean that it would depend on the concentration of sodium hydroxide solution. The pKa of ethanol is 15.9 which if you assume a 1M NaOH solution the pH would be about 14 (not including minute amounts of undisaccociated base). PH=pKa+log(A-/HA) -1.9=log(A-/HA) 0.0126 = (CH3CH2O-/CH3CH2OH). This means that the percent of ethoxide ions that are in solution is just over 1% (I can go into % ionization if you would like) which is certainly detectable. Similarly a 10M (difficult to achieve) solution of sodium hydroxide (pH=15) would have just over 10% ethoxide ions vs ethanol. The reaction occurs, but only under a weak equilibrium depending on concentration.
Yes
Its like
C2H5OH + NaOH = C2H5ONa + H2O
Yes
Its like
C2H5OH + NaOH = C2H5ONa + H2O
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i used to do experiments during my master. I used the weight measurement to make the solution. Let's say for 1l solution i.e. 1000ml. If you Want to make a 10% solution. Add 100 g of NaOH to 900ml (=900g) of water (De ionized). That will give you a 10% NaOH solution.
I hope this will help.
i used to do experiments during my master. I used the weight measurement to make the solution. Let's say for 1l solution i.e. 1000ml. If you Want to make a 10% solution. Add 100 g of NaOH to 900ml (=900g) of water (De ionized). That will give you a 10% NaOH solution.
I hope this will help.
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There is no "chemical reaction" but there ia a reaction in that the disassociation of the sodium and hydroxide ions and the hydration of those ions releases a LOT of heat, enough to boil water in some circumstances. Great care is needed.
There is no "chemical reaction" but there ia a reaction in that the disassociation of the sodium and hydroxide ions and the hydration of those ions releases a LOT of heat, enough to boil water in some circumstances. Great care is needed.
More
VOTE
Yes but everyone is forgetting that every reaction is an equilibrium. This would mean that it would depend on the concentration of sodium hydroxide solution. The pKa of ethanol is 15.9 which if you assume a 1M NaOH solution the pH would be about 14 (not including minute amounts of undisaccociated base). PH=pKa+log(A-/HA) -1.9=log(A-/HA) 0.0126 = (CH3CH2O-/CH3CH2OH). This means that the percent of ethoxide ions that are in solution is just over 1% (I can go into % ionization if you would like) which is certainly detectable. Similarly a 10M (difficult to achieve) solution of sodium hydroxide (pH=15) would have just over 10% ethoxide ions vs ethanol. The reaction occurs, but only under a weak equilibrium depending on concentration.
Yes but everyone is forgetting that every reaction is an equilibrium. This would mean that it would depend on the concentration of sodium hydroxide solution. The pKa of ethanol is 15.9 which if you assume a 1M NaOH solution the pH would be about 14 (not including minute amounts of undisaccociated base). PH=pKa+log(A-/HA) -1.9=log(A-/HA) 0.0126 = (CH3CH2O-/CH3CH2OH). This means that the percent of ethoxide ions that are in solution is just over 1% (I can go into % ionization if you would like) which is certainly detectable. Similarly a 10M (difficult to achieve) solution of sodium hydroxide (pH=15) would have just over 10% ethoxide ions vs ethanol. The reaction occurs, but only under a weak equilibrium depending on concentration.
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VOTE