When to use 1/Ka vs KbWhat software can calculate aqueous solution equilibria?Lewis base strength and the...
Past Perfect Tense
Examples of non trivial equivalence relations , I mean equivalence relations without the expression " same ... as" in their definition?
In gnome-terminal only 2 out of 3 zoom keys work
A question regarding using the definite article
When did stoichiometry begin to be taught in U.S. high schools?
What word means to make something obsolete?
Pressure to defend the relevance of one's area of mathematics
Did Henry V’s archers at Agincourt fight with no pants / breeches on because of dysentery?
Sci-fi novel series with instant travel between planets through gates. A river runs through the gates
How to determine the actual or "true" resolution of a digital photograph?
Does a creature that is immune to a condition still make a saving throw?
How to creep the reader out with what seems like a normal person?
Help, my Death Star suffers from Kessler syndrome!
Pulling the rope with one hand is as heavy as with two hands?
Colliding particles and Activation energy
How can I get precisely a certain cubic cm by changing the following factors?
Why do Ichisongas hate elephants and hippos?
What is the difference between `a[bc]d` (brackets) and `a{b,c}d` (braces)?
Feels like I am getting dragged in office politics
Why does Bran Stark feel that Jon Snow "needs to know" about his lineage?
Do I have an "anti-research" personality?
Where does the labelling of extrinsic semiconductors as "n" and "p" come from?
Why is the origin of “threshold” uncertain?
Is it cheaper to drop cargo drop than to land it?
When to use 1/Ka vs Kb
What software can calculate aqueous solution equilibria?Lewis base strength and the hydrogen protonpKa differences and reactivityAre my protolysis equations right?What is the relationship between the two definitions of the difference between basicity and nucleophilicity?Acid/Base dissociation constants relationshipWhy is water not part of the equilibrium constant?Combining acid dissociation constants to determine pH of diprotic acidWeird Wikipedia Section on Oxidizing Behavior of Nitric and Sulfuric AcidsQ. 36 2018 molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF
$begingroup$
I'm really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_mathrm{a}$ and $K_mathrm{b}$. For example here is a calculation where for the reaction involving $ce{NO2-}$, the equilibrium constant is 1 / $K_mathrm{a}$ even though it shows $ce{NO2-}$ acting as a base.
acid-base
edited 1 hour ago
Karsten Theis
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy 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 really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_mathrm{a}$ and $K_mathrm{b}$. For example here is a calculation where for the reaction involving $ce{NO2-}$, the equilibrium constant is 1 / $K_mathrm{a}$ even though it shows $ce{NO2-}$ acting as a base.
acid-base
edited 1 hour ago
Karsten Theis
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago
add a comment |
$begingroup$
I'm really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_mathrm{a}$ and $K_mathrm{b}$. For example here is a calculation where for the reaction involving $ce{NO2-}$, the equilibrium constant is 1 / $K_mathrm{a}$ even though it shows $ce{NO2-}$ acting as a base.
acid-base
edited 1 hour ago
Karsten Theis
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy 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 really unsure what the difference is. We're using these when calculating $K$ of combined reactions. In general, I am also confused what $K$ is in comparison to $K_mathrm{a}$ and $K_mathrm{b}$. For example here is a calculation where for the reaction involving $ce{NO2-}$, the equilibrium constant is 1 / $K_mathrm{a}$ even though it shows $ce{NO2-}$ acting as a base.
acid-base
acid-base
edited 1 hour ago
Karsten Theis
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Karsten Theis
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Karsten Theis
5,282644
edited 1 hour ago
Karsten Theis
5,282644
edited 1 hour ago
Karsten Theis
5,282644
5,282644
asked 2 hours ago
Lucky LucyLucky Lucy
111
New contributor
Lucky Lucy 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
Lucky LucyLucky Lucy
111
asked 2 hours ago
Lucky LucyLucky Lucy
111
111
New contributor
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lucky Lucy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago
add a comment |
1
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago
1
1
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrm{a}$, $K_mathrm{b}$, and $K_mathrm{w}$. If you reverse the first reaction, you get:
$$ce{NO2- + H3O+ <=> HNO2 + H2O} qquad K = 1/K_mathrm{a}tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_mathrm{a} times K_mathrm{b} = K_mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrm{a}$ vs $K_mathrm{b}$
Use the equilibrium constant that matches the reaction you are working on.
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
$endgroup$
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
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
});
}
});
Lucky Lucy is a new contributor. Be nice, and check out our Code of Conduct.
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%2f114517%2fwhen-to-use-1-ka-vs-kb%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
$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrm{a}$, $K_mathrm{b}$, and $K_mathrm{w}$. If you reverse the first reaction, you get:
$$ce{NO2- + H3O+ <=> HNO2 + H2O} qquad K = 1/K_mathrm{a}tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_mathrm{a} times K_mathrm{b} = K_mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrm{a}$ vs $K_mathrm{b}$
Use the equilibrium constant that matches the reaction you are working on.
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
$endgroup$
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
add a comment |
$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrm{a}$, $K_mathrm{b}$, and $K_mathrm{w}$. If you reverse the first reaction, you get:
$$ce{NO2- + H3O+ <=> HNO2 + H2O} qquad K = 1/K_mathrm{a}tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_mathrm{a} times K_mathrm{b} = K_mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrm{a}$ vs $K_mathrm{b}$
Use the equilibrium constant that matches the reaction you are working on.
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
$endgroup$
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
add a comment |
$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrm{a}$, $K_mathrm{b}$, and $K_mathrm{w}$. If you reverse the first reaction, you get:
$$ce{NO2- + H3O+ <=> HNO2 + H2O} qquad K = 1/K_mathrm{a}tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_mathrm{a} times K_mathrm{b} = K_mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrm{a}$ vs $K_mathrm{b}$
Use the equilibrium constant that matches the reaction you are working on.
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
$endgroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrm{a}$, $K_mathrm{b}$, and $K_mathrm{w}$. If you reverse the first reaction, you get:
$$ce{NO2- + H3O+ <=> HNO2 + H2O} qquad K = 1/K_mathrm{a}tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)
As you might know, the three equilibrium constants are related:
$$K_mathrm{a} times K_mathrm{b} = K_mathrm{w}$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrm{a}$ vs $K_mathrm{b}$
Use the equilibrium constant that matches the reaction you are working on.
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
edited 1 hour ago
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
answered 1 hour ago
Karsten TheisKarsten Theis
5,282644
5,282644
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
add a comment |
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
1
1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
$endgroup$
– Karsten Theis
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
$begingroup$
No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
add a comment |
Lucky Lucy is a new contributor. Be nice, and check out our Code of Conduct.
Lucky Lucy is a new contributor. Be nice, and check out our Code of Conduct.
Lucky Lucy is a new contributor. Be nice, and check out our Code of Conduct.
Lucky Lucy is a new contributor. Be nice, and check out our Code of Conduct.
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%2f114517%2fwhen-to-use-1-ka-vs-kb%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
1
$begingroup$
"We're using these when calculating k" — you do know what "these" (e.g. $K_mathrm{a}$ and $K_mathrm{b}$) are, and what $K$ refers to, don't you? As of now it's not quite clear to me what you are trying to illustrate with the example problem you provided.
$endgroup$
– andselisk
2 hours ago