Why is a weak base more able to deprotonate a strong acid than a weak acid?67 failures at extracting acetaminophen from Excedrin(R)Why is acetic acid more acidic than phenol?Acid/Base dissociation constants relationshipCan Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together?Why is sodium sulfate neutral while its conjugate acid is weak?Calculating the pH of buffer solution made of two salts of a polyprotic acidWhy do the amino groups in EDTA deprotonate first, in contrast to amino acids?How to justify this contradiction in acid-base equilibrium of methanol and water?Why do most carboxylic acids have high pKa (~5) in spite of having a conjugate base ion that is stabilized by resonance?Calculate pH of a weak acid and weak base neutralization reaction
How is TD(0) method helpful? What good does it do?
Is it possible for people to live in the eye of a permanent hypercane?
Are the AT-AT's from "Empire Strikes Back" a deliberate reference to Mecha?
My coworkers think I had a long honeymoon. Actually I was diagnosed with cancer. How do I talk about it?
How to pass a regex when finding a directory path in bash?
What is the traditional way of earning a doctorate in Germany?
What are the words for people who cause trouble believing they know better?
What is the right way to float a home lab?
Secure offsite backup, even in the case of hacker root access
What risks are there when you clear your cookies instead of logging off?
What is the advantage of carrying a tripod and ND-filters when you could use image stacking instead?
What's the correct term describing the action of sending a brand-new ship out into its first seafaring trip?
How do photons get into the eyes?
Did Darth Vader wear the same suit for 20+ years?
Did thousands of women die every year due to illegal abortions before Roe v. Wade?
Company did not petition for visa in a timely manner. Is asking me to work from overseas, but wants me to take a paycut
What happens if you do emergency landing on a US base in middle of the ocean?
How can Iron Man's suit withstand this?
Building a road to escape Earth's gravity by making a pyramid on Antartica
Movie where a boy is transported into the future by an alien spaceship
Bent spoke design wheels — feasible?
Payment instructions from HomeAway look fishy to me
How could a government be implemented in a virtual reality?
Avoiding cliches when writing gods
Why is a weak base more able to deprotonate a strong acid than a weak acid?
67 failures at extracting acetaminophen from Excedrin(R)Why is acetic acid more acidic than phenol?Acid/Base dissociation constants relationshipCan Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together?Why is sodium sulfate neutral while its conjugate acid is weak?Calculating the pH of buffer solution made of two salts of a polyprotic acidWhy do the amino groups in EDTA deprotonate first, in contrast to amino acids?How to justify this contradiction in acid-base equilibrium of methanol and water?Why do most carboxylic acids have high pKa (~5) in spite of having a conjugate base ion that is stabilized by resonance?Calculate pH of a weak acid and weak base neutralization reaction
$begingroup$
Very basic question, but I'm rather confused.
Say I want to separate a carboxylic acid $(ceRCOOH,$ $mathrmpK_mathrma~5)$ from a phenol $(ceArOH,$ $mathrmpK_mathrma~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(ceNaHCO3),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).
The question is:
Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?
If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?
acid-base extraction
asked May 19 at 16:08
chompionchompion
796
$endgroup$
add a comment |
$begingroup$
Very basic question, but I'm rather confused.
Say I want to separate a carboxylic acid $(ceRCOOH,$ $mathrmpK_mathrma~5)$ from a phenol $(ceArOH,$ $mathrmpK_mathrma~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(ceNaHCO3),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).
The question is:
Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?
If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?
acid-base extraction
asked May 19 at 16:08
chompionchompion
796
$endgroup$
3
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
1
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29
add a comment |
$begingroup$
Very basic question, but I'm rather confused.
Say I want to separate a carboxylic acid $(ceRCOOH,$ $mathrmpK_mathrma~5)$ from a phenol $(ceArOH,$ $mathrmpK_mathrma~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(ceNaHCO3),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).
The question is:
Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?
If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?
acid-base extraction
asked May 19 at 16:08
chompionchompion
796
$endgroup$
Very basic question, but I'm rather confused.
Say I want to separate a carboxylic acid $(ceRCOOH,$ $mathrmpK_mathrma~5)$ from a phenol $(ceArOH,$ $mathrmpK_mathrma~10)$ via acid-base extraction. Apparently I am able to use sodium bicarbonate $(ceNaHCO3),$ a weak base, as a way to deprotonate the stronger acid, carboxylic acid. This allows the carboxylic acid to dissolve into the aqueous phase in an ether extraction, as shown in the far left of the attached diagram, while the phenol would remain in the organic phase (A at the bottom corresponds to the carboxylic acid, B is the phenol, C is the amine).
The question is:
Why does a weak base 'prefer' to deprotonate a stronger acid? Is it because, being a stronger acid, the carboxylic acid will have more anions in solution compared to the weaker acid, so the (relatively few) cations of the weak base will be more likely to find carboxylates than phenoxides?
If this is the case, could we still use a strong base, but just in a very very low concentration and have the same effect as using a weak base at a higher concentration?
acid-base extraction
acid-base extraction
asked May 19 at 16:08
chompionchompion
796
asked May 19 at 16:08
chompionchompion
796
edited May 19 at 16:34
chompion
asked May 19 at 16:08
chompionchompion
796
asked May 19 at 16:08
chompionchompion
796
asked May 19 at 16:08
chompionchompion
796
796
3
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
1
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29
add a comment |
3
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
1
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29
3
3
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
1
1
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $ceCO2$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $ceNaOH$ (a strong base) or $ceNaHCO3$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $ceOH-$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $ceRCOOH$ and $ceHCO3-$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.
answered May 19 at 16:34
TAR86TAR86
5,16711033
$endgroup$
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "431"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f115568%2fwhy-is-a-weak-base-more-able-to-deprotonate-a-strong-acid-than-a-weak-acid%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $ceCO2$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $ceNaOH$ (a strong base) or $ceNaHCO3$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $ceOH-$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $ceRCOOH$ and $ceHCO3-$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.
answered May 19 at 16:34
TAR86TAR86
5,16711033
$endgroup$
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
add a comment |
$begingroup$
The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $ceCO2$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $ceNaOH$ (a strong base) or $ceNaHCO3$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $ceOH-$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $ceRCOOH$ and $ceHCO3-$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.
answered May 19 at 16:34
TAR86TAR86
5,16711033
$endgroup$
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
add a comment |
$begingroup$
The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $ceCO2$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $ceNaOH$ (a strong base) or $ceNaHCO3$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $ceOH-$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $ceRCOOH$ and $ceHCO3-$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.
answered May 19 at 16:34
TAR86TAR86
5,16711033
$endgroup$
The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $ceCO2$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $ceNaOH$ (a strong base) or $ceNaHCO3$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $ceOH-$, because any stronger base will just deprotonate a readily available water molecule.
To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $ceRCOOH$ and $ceHCO3-$ do not have to come close to each other to effectively transfer a proton between them.
Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.
answered May 19 at 16:34
TAR86TAR86
5,16711033
answered May 19 at 16:34
TAR86TAR86
5,16711033
answered May 19 at 16:34
TAR86TAR86
5,16711033
answered May 19 at 16:34
TAR86TAR86
5,16711033
5,16711033
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
add a comment |
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
$begingroup$
I see... so it's similar to electrons jumping from atom to atom through a solid wire, but in this case it's protons moving in all directions in a fluid, rather than a solid. I've read strong acids and bases are better conductors of electricity. As the protons jump from the 'better conductor,' because it deprotonates more easily, the stronger acid becomes negatively charged and separates into the aqueous solution. Correct...?
$endgroup$
– chompion
May 19 at 16:50
1
1
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
$begingroup$
@chompion That's basically correct.
$endgroup$
– TAR86
May 19 at 17:06
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f115568%2fwhy-is-a-weak-base-more-able-to-deprotonate-a-strong-acid-than-a-weak-acid%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
The picture is a bit confusing: Phenol should $cePhOH$ and nothing else. Sometimes, $ceAr$ is used to denote an aromatic compound, but that does not seem to be the case here.
$endgroup$
– TAR86
May 19 at 16:17
1
$begingroup$
@TAR86, agreed, I've updated the diagram with a full screenshot including the question and solution, in case it helps any. In any case I believe they just want to get the point across that a weak base can be used to extract a strong acid, and conceptually I'm trying to understand why.
$endgroup$
– chompion
May 19 at 16:29