Which acid/base does a strong base/acid react when added to a buffer solution? The Next CEO of Stack OverflowHow to calculate the pH of a buffered solution with Henderson Hasselbalch?What's happening at the beginning of a weak acid titration?Buffer and strong acidWhy is only one part of buffer in the reactants side?What is causing the buffer region in a weak acid - strong base titration?Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?Buffer Solution Problem (Acid/Conjugate Base)Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base?How is preparing a buffer possible?Error with explaination of shape of weak acid strong base titration?
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Which acid/base does a strong base/acid react when added to a buffer solution?
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Which acid/base does a strong base/acid react when added to a buffer solution?
The Next CEO of Stack OverflowHow to calculate the pH of a buffered solution with Henderson Hasselbalch?What's happening at the beginning of a weak acid titration?Buffer and strong acidWhy is only one part of buffer in the reactants side?What is causing the buffer region in a weak acid - strong base titration?Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?Buffer Solution Problem (Acid/Conjugate Base)Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base?How is preparing a buffer possible?Error with explaination of shape of weak acid strong base titration?
$begingroup$
I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $ceCH3COOH$ and $ceNaCH3COO$ with $K_a = 1.8 times 10^-5$, we might want to find what the pH will become if $ce0.010 M KOH$ is added. Given the dissociation of acetic acid:
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid ($ceCH3COOH$ or the conjugate $ceH3O+$) will the base $ceKOH$ react with? My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the forward direction, but why wouldn't it react with $ceCH3COOH$ since it also is an acid?
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
acid-base equilibrium
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $ceCH3COOH$ and $ceNaCH3COO$ with $K_a = 1.8 times 10^-5$, we might want to find what the pH will become if $ce0.010 M KOH$ is added. Given the dissociation of acetic acid:
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid ($ceCH3COOH$ or the conjugate $ceH3O+$) will the base $ceKOH$ react with? My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the forward direction, but why wouldn't it react with $ceCH3COOH$ since it also is an acid?
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
acid-base equilibrium
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago
add a comment |
$begingroup$
I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $ceCH3COOH$ and $ceNaCH3COO$ with $K_a = 1.8 times 10^-5$, we might want to find what the pH will become if $ce0.010 M KOH$ is added. Given the dissociation of acetic acid:
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid ($ceCH3COOH$ or the conjugate $ceH3O+$) will the base $ceKOH$ react with? My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the forward direction, but why wouldn't it react with $ceCH3COOH$ since it also is an acid?
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
acid-base equilibrium
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of $ceCH3COOH$ and $ceNaCH3COO$ with $K_a = 1.8 times 10^-5$, we might want to find what the pH will become if $ce0.010 M KOH$ is added. Given the dissociation of acetic acid:
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid ($ceCH3COOH$ or the conjugate $ceH3O+$) will the base $ceKOH$ react with? My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the forward direction, but why wouldn't it react with $ceCH3COOH$ since it also is an acid?
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
acid-base equilibrium
acid-base equilibrium
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 mins ago
Andrew Li
asked 2 hours ago
Andrew LiAndrew Li
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Andrew LiAndrew Li
1084
asked 2 hours ago
Andrew LiAndrew Li
1084
1084
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andrew Li is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago
add a comment |
$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago
$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$ceH3O+ + OH- <=> 2 H2O$$
If you add up the two reaction, you get a third one:
$$ceCH3COOH + OH- <=> H2O + CH3COO-$$
You can go either way (use the second or the third reaction) to explain that the concentration of $ceOH-$ drops and the concentration of $ceCH3COOH$ increases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $pue-5$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
$endgroup$
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
add a comment |
$begingroup$
The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.
answered 1 hour ago
MaxWMaxW
15.2k22261
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$ceH3O+ + OH- <=> 2 H2O$$
If you add up the two reaction, you get a third one:
$$ceCH3COOH + OH- <=> H2O + CH3COO-$$
You can go either way (use the second or the third reaction) to explain that the concentration of $ceOH-$ drops and the concentration of $ceCH3COOH$ increases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $pue-5$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
$endgroup$
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
add a comment |
$begingroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$ceH3O+ + OH- <=> 2 H2O$$
If you add up the two reaction, you get a third one:
$$ceCH3COOH + OH- <=> H2O + CH3COO-$$
You can go either way (use the second or the third reaction) to explain that the concentration of $ceOH-$ drops and the concentration of $ceCH3COOH$ increases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $pue-5$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
$endgroup$
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
add a comment |
$begingroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$ceH3O+ + OH- <=> 2 H2O$$
If you add up the two reaction, you get a third one:
$$ceCH3COOH + OH- <=> H2O + CH3COO-$$
You can go either way (use the second or the third reaction) to explain that the concentration of $ceOH-$ drops and the concentration of $ceCH3COOH$ increases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $pue-5$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
$endgroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$ceH3O+ + OH- <=> 2 H2O$$
If you add up the two reaction, you get a third one:
$$ceCH3COOH + OH- <=> H2O + CH3COO-$$
You can go either way (use the second or the third reaction) to explain that the concentration of $ceOH-$ drops and the concentration of $ceCH3COOH$ increases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $pue-5$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
edited 1 hour ago
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
answered 1 hour ago
Karsten TheisKarsten Theis
3,687541
3,687541
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
add a comment |
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
"that the concentration of $~ce OH−$ $colorredtextdrops$ and the concentration of $~ce CH3COOH$ $colorredtextincreases$."I think:that the concentration of $ce OH−$ $colorgreentextincreases$ and the concentration of$ce CH3COOH$ $colorgreentextdecreases$.
$endgroup$
– Adnan AL-Amleh
51 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
$begingroup$
Thanks for the comprehensive answer. One question: you said that the direction I choose shouldn't affect pH. But in the Henderson-Hasselbalch equation don't you have to compute $-log([base]/[acid])$? If I choose different directions, won't the new concentrations of base and acid be different?
$endgroup$
– Andrew Li
7 mins ago
add a comment |
$begingroup$
The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
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$begingroup$
The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The hydroxide ion reacts with the hydronium ion (I hate that name; I prefer the simpler "hydrogen ion") to increase the concentration of water and decrease the concentration of hydrogen ion. To restore the equilibrium, now more acetic acid will ionize, shifting the equilibrium to the right as you have it written above.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
answered 2 hours ago
Charlie AmelottiCharlie Amelotti
94
94
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Charlie Amelotti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.
answered 1 hour ago
MaxWMaxW
15.2k22261
$endgroup$
add a comment |
$begingroup$
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.
answered 1 hour ago
MaxWMaxW
15.2k22261
$endgroup$
add a comment |
$begingroup$
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.
answered 1 hour ago
MaxWMaxW
15.2k22261
$endgroup$
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a solution very conductive. So just based on chance, a hydroxide ion is more likely to reaction with a hydronium ion than an acetic acid molecule. Hence your teacher's remark.
However Zhe is correct in that it really doesn't make a difference overall. The H+/acetate/acetic acid equilibrium will be established "fairly" quickly.
answered 1 hour ago
MaxWMaxW
15.2k22261
answered 1 hour ago
MaxWMaxW
15.2k22261
answered 1 hour ago
MaxWMaxW
15.2k22261
answered 1 hour ago
MaxWMaxW
15.2k22261
15.2k22261
add a comment |
add a comment |
Andrew Li is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Li is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Li is a new contributor. Be nice, and check out our Code of Conduct.
Andrew Li is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
They're all in equilibrium together. It would be mistake to ask which one reacted. Whatever you pick is still part of the equilibrium that needs to readjust.
$endgroup$
– Zhe
2 hours ago
$begingroup$
"My teacher said that the base will react with $ceH3O+$ to shift equilibrium in the $colorredreverse$ direction,".I think to shift the equilibrium in the $colorgreenforward$ direction.
$endgroup$
– Adnan AL-Amleh
1 hour ago