When to use 1/Ka vs Kb
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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
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asked 2 hours ago
Lucky LucyLucky Lucy
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add a comment |
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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
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$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
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
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1
$begingroup$
@andselisk - Thanks for aligning and adding subscripts.
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– Karsten Theis
1 hour ago
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No prob, Mathew did most of the editing, I only polished the formatting a bit more:)
$endgroup$
– andselisk
1 hour ago
add a comment |
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$begingroup$
$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$
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 |
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$begingroup$
"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