Which acid/base does a strong base/acid react when added to a buffer solution? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How 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?
How do I keep my slimes from escaping their pens?
Compare a given version number in the form major.minor.build.patch and see if one is less than the other
Do I really need recursive chmod to restrict access to a folder?
Generate an RGB colour grid
How to remove list items depending on predecessor in python
Why aren't air breathing engines used as small first stages
A binary hook-length formula?
Why do we bend a book to keep it straight?
Can I cast Passwall to drop an enemy into a 20-foot pit?
When were vectors invented?
Should I discuss the type of campaign with my players?
What's the meaning of 間時肆拾貳 at a car parking sign
Is it a good idea to use CNN to classify 1D signal?
What is Wonderstone and are there any references to it pre-1982?
Is it true that "carbohydrates are of no use for the basal metabolic need"?
Seeking colloquialism for “just because”
Can we see the USA flag on the Moon from Earth?
How to find all the available tools in mac terminal?
List of Python versions
Remainder theorem for polynomials (JUEE 1990)
Ring Automorphisms that fix 1.
Why did the US and UK choose different solutions to the problem of an undemocratic upper house?
Why are both D and D# fitting into my E minor key?
Why did the IBM 650 use bi-quinary?
Which acid/base does a strong base/acid react when added to a buffer solution?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How 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 Apr 1 at 19:28
Andrew LiAndrew Li
1205
$endgroup$
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 Apr 1 at 19:28
Andrew LiAndrew Li
1205
$endgroup$
$begingroup$
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
Apr 1 at 19:41
$begingroup$
"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
Apr 1 at 20:34
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 Apr 1 at 19:28
Andrew LiAndrew Li
1205
$endgroup$
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 Apr 1 at 19:28
Andrew LiAndrew Li
1205
asked Apr 1 at 19:28
Andrew LiAndrew Li
1205
edited Apr 1 at 21:37
Andrew Li
asked Apr 1 at 19:28
Andrew LiAndrew Li
1205
asked Apr 1 at 19:28
Andrew LiAndrew Li
1205
asked Apr 1 at 19:28
Andrew LiAndrew Li
1205
1205
$begingroup$
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
Apr 1 at 19:41
$begingroup$
"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
Apr 1 at 20:34
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
Apr 1 at 19:41
$begingroup$
"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
Apr 1 at 20:34
$begingroup$
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
Apr 1 at 19:41
$begingroup$
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
Apr 1 at 19:41
$begingroup$
"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
Apr 1 at 20:34
$begingroup$
"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
Apr 1 at 20:34
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 $ceOH-$ is part of a neutralization reaction. As a consequence, the concentration of $ceCH3COO-$ increases and that of $ceCH3COOH$ and hydronium decreases.
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 Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
$endgroup$
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
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.
$endgroup$
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 Apr 1 at 19:59
MaxWMaxW
15.8k22261
$endgroup$
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%2f111957%2fwhich-acid-base-does-a-strong-base-acid-react-when-added-to-a-buffer-solution%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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 $ceOH-$ is part of a neutralization reaction. As a consequence, the concentration of $ceCH3COO-$ increases and that of $ceCH3COOH$ and hydronium decreases.
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 Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
$endgroup$
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
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 $ceOH-$ is part of a neutralization reaction. As a consequence, the concentration of $ceCH3COO-$ increases and that of $ceCH3COOH$ and hydronium decreases.
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 Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
$endgroup$
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
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 $ceOH-$ is part of a neutralization reaction. As a consequence, the concentration of $ceCH3COO-$ increases and that of $ceCH3COOH$ and hydronium decreases.
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 Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
$endgroup$
$$ceCH3COOH + H2O <=> H3O+ + CH3COO-$$
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 $ceOH-$ is part of a neutralization reaction. As a consequence, the concentration of $ceCH3COO-$ increases and that of $ceCH3COOH$ and hydronium decreases.
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 Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
edited Apr 1 at 21:58
answered Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
answered Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
answered Apr 1 at 20:11
Karsten TheisKarsten Theis
4,624542
4,624542
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
"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
Apr 1 at 20:59
$begingroup$
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
Apr 1 at 21:44
$begingroup$
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
Apr 1 at 21:44
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Adnan AL-Amleh I fixed my typo: The neutralization reaction results in a drop of hydroxide and weak acid, while the concentration of weak base increases.
$endgroup$
– Karsten Theis
Apr 1 at 21:55
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
$begingroup$
@Andrew_Li I had a typo which I fixed, and I revised my answer to address your follow-up question. No matter which reaction you start with, the ratio of weak base to weak acid concentration will increase, so the pH will increase. You expect that because that is what strong bases do. Because the solution is buffered, the change is much smaller than it would be for pure water, but it still changes a bit in that direction.
$endgroup$
– Karsten Theis
Apr 1 at 22:01
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 Apr 1 at 19:47
user76478
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 Apr 1 at 19:59
MaxWMaxW
15.8k22261
$endgroup$
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 Apr 1 at 19:59
MaxWMaxW
15.8k22261
$endgroup$
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 Apr 1 at 19:59
MaxWMaxW
15.8k22261
$endgroup$
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 Apr 1 at 19:59
MaxWMaxW
15.8k22261
answered Apr 1 at 19:59
MaxWMaxW
15.8k22261
answered Apr 1 at 19:59
MaxWMaxW
15.8k22261
answered Apr 1 at 19:59
MaxWMaxW
15.8k22261
15.8k22261
add a comment |
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%2f111957%2fwhich-acid-base-does-a-strong-base-acid-react-when-added-to-a-buffer-solution%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
$begingroup$
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
Apr 1 at 19:41
$begingroup$
"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
Apr 1 at 20:34