Acid base a2
Q. Why are solutions of sodium ethanoate slightly alkaline?
A. The sodium ions react with water to give an alkali
B. The ethanoate ions react with water to give hydroxide ions
C. All sodium salts give alkaline solutions
D. The sodium ethanoate is fully ionized in solution
The answer is B but why is not A!!!!!!!
The reaction that occurs is CH3COONa + H2O <=> CH3COOH + NaOH
Sodium ions reacted with water to give an alkali??!!!! Plus ethanoate ions did noy give hydroxide ions... Someone pls explain 😢😭
A. The sodium ions react with water to give an alkali
B. The ethanoate ions react with water to give hydroxide ions
C. All sodium salts give alkaline solutions
D. The sodium ethanoate is fully ionized in solution
The answer is B but why is not A!!!!!!!
The reaction that occurs is CH3COONa + H2O <=> CH3COOH + NaOH
Sodium ions reacted with water to give an alkali??!!!! Plus ethanoate ions did noy give hydroxide ions... Someone pls explain 😢😭
Original post by pondsteps
Q. Why are solutions of sodium ethanoate slightly alkaline?
A. The sodium ions react with water to give an alkali
B. The ethanoate ions react with water to give hydroxide ions
C. All sodium salts give alkaline solutions
D. The sodium ethanoate is fully ionized in solution
The answer is B but why is not A!!!!!!!
The reaction that occurs is CH3COONa + H2O <=> CH3COOH + NaOH
Sodium ions reacted with water to give an alkali??!!!! Plus ethanoate ions did noy give hydroxide ions... Someone pls explain 😢😭
A. The sodium ions react with water to give an alkali
B. The ethanoate ions react with water to give hydroxide ions
C. All sodium salts give alkaline solutions
D. The sodium ethanoate is fully ionized in solution
The answer is B but why is not A!!!!!!!
The reaction that occurs is CH3COONa + H2O <=> CH3COOH + NaOH
Sodium ions reacted with water to give an alkali??!!!! Plus ethanoate ions did noy give hydroxide ions... Someone pls explain 😢😭
Sodium ions are unreactive in water, Sodium atoms react rapidly leaving an alkaline solution and evolving H2(g). So the statement in A is simply untrue.
It must be B because ethanoate is the conjugate base of a weak acid (ethanoic acid) and so an equilibrium will be established in which some of the ethanoate has become protonated and some of the water has become deprotonated.
Deprotonated water is simply OH- so the solution is slightly alkaline
(edited 7 years ago)
Original post by MexicanKeith
Sodium ions are unreactive in water, Sodium atoms react rapidly leaving an alkaline solution and evolving H2(g). So the statement in A is simply untrue.
It must be B because ethanoate is the conjugate base of a weak acid (ethanoic acid) and so an equilibrium will be established in which some of the ethanoate has become protonated and some of the water has become deprotonated.
Deprotonated water is simply OH- so the solution is slightly alkaline
It must be B because ethanoate is the conjugate base of a weak acid (ethanoic acid) and so an equilibrium will be established in which some of the ethanoate has become protonated and some of the water has become deprotonated.
Deprotonated water is simply OH- so the solution is slightly alkaline
Ohh... So sodium reacts covalently with water???? Why cant Na+ + OH- --> NaOH?????
Original post by pondsteps
Ohh... So sodium reacts covalently with water???? Why cant Na+ + OH- --> NaOH?????
The answer is in the wording of the answers: (A) states that the sodium ions react with the OH- ions. This is not really true in regards to why this reaction happens, this bit just happens at the end.
(B) on the other hand states that the ethanoate ions react with water to form hydroxide ions. This is correct in its entirety. This is quite a difficult little answer to explain but I hope that helps.
Alex
(edited 7 years ago)
Original post by ajsullivan
NO It doesnt react with water covalently. Going back to the questions, the number of OH- ions in water is negligable (virtually zero). This is because water hardly splits up into H+ + OH- . When the sodium ethanoate is put into water is totally dissociates into CH3COO- and Na+ as it is an ionic salt (that is to say in water there is now zero CH3COONa). ONCE the sodium ethanoate dissociates, then the CH3COO- ions that have been produced would pair up with the H+ ions that are available, and hence would leave some OH- ions.
The answer is in the wording of the answers: (A) states that the sodium ions react with the OH- ions. This is not really true in regards to why this reaction happens, this bit just happens at the end.
(B) on the other hand states that the ethanoate ions react with water to form hydroxide ions. This is correct in its entirety. This is quite a difficult little answer to explain but I hope that helps.
Alex
The answer is in the wording of the answers: (A) states that the sodium ions react with the OH- ions. This is not really true in regards to why this reaction happens, this bit just happens at the end.
(B) on the other hand states that the ethanoate ions react with water to form hydroxide ions. This is correct in its entirety. This is quite a difficult little answer to explain but I hope that helps.
Alex
i have a tiny question off topic.... if sodium ions cant react then how does double decomposition occur between salts if they have a stable configuration !!
another q .. i dont get the idea of the buffer 100%
like extra ethanoate ions are being formed then what about H+.. what if they run out ?? then only excess of ethanote ions remain....... so whats the pointI don't think anyone said sodium ions can't react. The previous point was that it doesn't react in the above like you thought. It was the ethanoate reacting with the water and then at the end the sodium ions pairing up with the OH- ions.
Your question about double decomposition reactions is best off answered with a generalised answer as I imagine it would massively depend on the actual reaction you are referring to. Reactions occur in the way they do because the products created are energetically more favourable than the ones they started as. If that means to just swap metal ions, then so be it. WHY they are more energetically more stable is a very complicated question.
Your question about buffers sounds like you need to do a little bit of in depth revision on the topic. This topic is on my website but is not yet written into html (its my next one actually!), but it is in PDF format, I suggest you check it out.... Go to my website Science Skool and click on A2 physical chemistry and then Acids, Bases and Buffers.
Let me know how it goes
Alex
Your question about double decomposition reactions is best off answered with a generalised answer as I imagine it would massively depend on the actual reaction you are referring to. Reactions occur in the way they do because the products created are energetically more favourable than the ones they started as. If that means to just swap metal ions, then so be it. WHY they are more energetically more stable is a very complicated question.
Your question about buffers sounds like you need to do a little bit of in depth revision on the topic. This topic is on my website but is not yet written into html (its my next one actually!), but it is in PDF format, I suggest you check it out.... Go to my website Science Skool and click on A2 physical chemistry and then Acids, Bases and Buffers.
Let me know how it goes
Alex
Original post by ajsullivan
I don't think anyone said sodium ions can't react. The previous point was that it doesn't react in the above like you thought. It was the ethanoate reacting with the water and then at the end the sodium ions pairing up with the OH- ions.
Your question about double decomposition reactions is best off answered with a generalised answer as I imagine it would massively depend on the actual reaction you are referring to. Reactions occur in the way they do because the products created are energetically more favourable than the ones they started as. If that means to just swap metal ions, then so be it. WHY they are more energetically more stable is a very complicated question.
Your question about buffers sounds like you need to do a little bit of in depth revision on the topic. This topic is on my website but is not yet written into html (its my next one actually!), but it is in PDF format, I suggest you check it out.... Go to my website Science Skool and click on A2 physical chemistry and then Acids, Bases and Buffers.
Let me know how it goes
Alex
Your question about double decomposition reactions is best off answered with a generalised answer as I imagine it would massively depend on the actual reaction you are referring to. Reactions occur in the way they do because the products created are energetically more favourable than the ones they started as. If that means to just swap metal ions, then so be it. WHY they are more energetically more stable is a very complicated question.
Your question about buffers sounds like you need to do a little bit of in depth revision on the topic. This topic is on my website but is not yet written into html (its my next one actually!), but it is in PDF format, I suggest you check it out.... Go to my website Science Skool and click on A2 physical chemistry and then Acids, Bases and Buffers.
Let me know how it goes
Alex
Omg this website has amazzzzzing notes i'll check out the other topics soon as well, thanks soooo much!!
As for my first question, can you pls read the response above your one and you'll understand what im talking about
CH3COONa is ionic: it is composed of CH3COO- ions as well as Na+ ions. The negative charge on the ethanoate anion (think in terms of curly arrows) can remove a proton from a molecule of water, leaving behind an OH- ion.
Remember we have Na+ ions floating around in solution, so we form an NaOH solution, ie. an alkaline solution.
Remember we have Na+ ions floating around in solution, so we form an NaOH solution, ie. an alkaline solution.
Original post by ironmaidenfan
CH3COONa is ionic: it is composed of CH3COO- ions as well as Na+ ions. The negative charge on the ethanoate anion (think in terms of curly arrows) can remove a proton from a molecule of water, leaving behind an OH- ion.
Remember we have Na+ ions floating around in solution, so we form an NaOH solution, ie. an alkaline solution.
Remember we have Na+ ions floating around in solution, so we form an NaOH solution, ie. an alkaline solution.
you mean that A is also correct?
I am glad you like the site, it is a work in progress. Pass it on to your friends.
As for the previous comment he said that sodium ions are unreactive in water and I think he was getting at the fact that if the sodium is an ION it has already lost its electron and has a positive charge. To do this it must have given this electron to something else (in your example the ethanoate ion). When a double decomposition occurs you can think of the ions as swapping over, such as:
Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3
There is no oxidation state change here, no loss or gain of electrons, just swapping. It happens because it must be energetically more stable to be this way otherwise it would not be feasible/spontaneous. TO go into more detail here you would need to understand entropy. There is a clear decrease in entropy here so the fact that my example does happen spontaneously tells me this is exothermic. This would then lead on to Gibbs Free Energy calculations.
Alex
As for the previous comment he said that sodium ions are unreactive in water and I think he was getting at the fact that if the sodium is an ION it has already lost its electron and has a positive charge. To do this it must have given this electron to something else (in your example the ethanoate ion). When a double decomposition occurs you can think of the ions as swapping over, such as:
Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3
There is no oxidation state change here, no loss or gain of electrons, just swapping. It happens because it must be energetically more stable to be this way otherwise it would not be feasible/spontaneous. TO go into more detail here you would need to understand entropy. There is a clear decrease in entropy here so the fact that my example does happen spontaneously tells me this is exothermic. This would then lead on to Gibbs Free Energy calculations.
Alex
Original post by pondsteps
you mean that A is also correct?
I'm just going to try to clear up what's happening here, hopefully you'll see that the sodium ions just stay as sodium ions all the way through
First we put sodium ethanoate into water, this is an ionic compound, soluble in water, so it exists in solution as a sodium ion and an ethanoate ion.
CH3COONa(s) ----> CH3COO-(aq) + Na+(aq)
In this situation the sodium ion is quite stable, it is wrapped up in a layer of solvent water and it doesn't react with water.
Ethanoate however, can gain a proton from water to become ethanoic acid, so the ethanoate enters equilibirum with ethanoic acid (it does this because it is a weak acid so readily exchanges protons with water)
CH3COO-(aq) + H2O(l) <----> CH3COOH(aq) + OH-(aq)
When we first added the ethanoate, there was no ethanoic acid present, in the process of reaching equilibrium, the concentration of reactants falls and the concentration of product rises.
This mean the concentration OH-(aq) has risen and the product is alkaline!
I think you might be getting confused with the fact that the reaction taking place can also be written as
CH3COONa(aq) + H2O(l) <-----> CH3COOH(aq) + NaOH(aq)
This is just the same as the previous equation I wrote but now the sodium ions are included! The confusion is arising because you see NaOH is formed and think that the sodium has reacted to form sodium hydroxide.
But remember, Sodium hydroxide is an ionic compound, very soluble in water, so it exists as ions (just like sodium ethanoate)!
If you think about what happens to the sodium through the reaction, it starts out as Na+(aq) stable and wrapped up in water and it finishes up as Na+(aq) still stable and wrapped up in water!
In other words it plays no part in the reaction! it is what you would call a spectator ion.
So the NaOH(aq) arises because the ethanoate reacts with water leaving OH-(aq) meanwhile the sodium ion sits in solution, not doing anything!
Hope that helps
Original post by MexicanKeith
I'm just going to try to clear up what's happening here, hopefully you'll see that the sodium ions just stay as sodium ions all the way through
First we put sodium ethanoate into water, this is an ionic compound, soluble in water, so it exists in solution as a sodium ion and an ethanoate ion.
CH3COONa(s) ----> CH3COO-(aq) + Na+(aq)
In this situation the sodium ion is quite stable, it is wrapped up in a layer of solvent water and it doesn't react with water.
Ethanoate however, can gain a proton from water to become ethanoic acid, so the ethanoate enters equilibirum with ethanoic acid (it does this because it is a weak acid so readily exchanges protons with water)
CH3COO-(aq) + H2O(l) <----> CH3COOH(aq) + OH-(aq)
When we first added the ethanoate, there was no ethanoic acid present, in the process of reaching equilibrium, the concentration of reactants falls and the concentration of product rises.
This mean the concentration OH-(aq) has risen and the product is alkaline!
I think you might be getting confused with the fact that the reaction taking place can also be written as
CH3COONa(aq) + H2O(l) <-----> CH3COOH(aq) + NaOH(aq)
This is just the same as the previous equation I wrote but now the sodium ions are included! The confusion is arising because you see NaOH is formed and think that the sodium has reacted to form sodium hydroxide.
But remember, Sodium hydroxide is an ionic compound, very soluble in water, so it exists as ions (just like sodium ethanoate)!
If you think about what happens to the sodium through the reaction, it starts out as Na+(aq) stable and wrapped up in water and it finishes up as Na+(aq) still stable and wrapped up in water!
In other words it plays no part in the reaction! it is what you would call a spectator ion.
So the NaOH(aq) arises because the ethanoate reacts with water leaving OH-(aq) meanwhile the sodium ion sits in solution, not doing anything!
Hope that helps
First we put sodium ethanoate into water, this is an ionic compound, soluble in water, so it exists in solution as a sodium ion and an ethanoate ion.
CH3COONa(s) ----> CH3COO-(aq) + Na+(aq)
In this situation the sodium ion is quite stable, it is wrapped up in a layer of solvent water and it doesn't react with water.
Ethanoate however, can gain a proton from water to become ethanoic acid, so the ethanoate enters equilibirum with ethanoic acid (it does this because it is a weak acid so readily exchanges protons with water)
CH3COO-(aq) + H2O(l) <----> CH3COOH(aq) + OH-(aq)
When we first added the ethanoate, there was no ethanoic acid present, in the process of reaching equilibrium, the concentration of reactants falls and the concentration of product rises.
This mean the concentration OH-(aq) has risen and the product is alkaline!
I think you might be getting confused with the fact that the reaction taking place can also be written as
CH3COONa(aq) + H2O(l) <-----> CH3COOH(aq) + NaOH(aq)
This is just the same as the previous equation I wrote but now the sodium ions are included! The confusion is arising because you see NaOH is formed and think that the sodium has reacted to form sodium hydroxide.
But remember, Sodium hydroxide is an ionic compound, very soluble in water, so it exists as ions (just like sodium ethanoate)!
If you think about what happens to the sodium through the reaction, it starts out as Na+(aq) stable and wrapped up in water and it finishes up as Na+(aq) still stable and wrapped up in water!
In other words it plays no part in the reaction! it is what you would call a spectator ion.
So the NaOH(aq) arises because the ethanoate reacts with water leaving OH-(aq) meanwhile the sodium ion sits in solution, not doing anything!
Hope that helps
So you mean there is no actual rxn (exchange of electrons) they are only attracted to each other
Original post by pondsteps
So you mean there is no actual rxn (exchange of electrons) they are only attracted to each other
There is no actual reaction involving the Sodium ions
Original post by MexicanKeith
There is no actual reaction involving the Sodium ions
Okay thank you very much xxx
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