When to use 1/Ka vs Kb












1












$begingroup$


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.



enter image description here






acid-base







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  • 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












$begingroup$


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.



enter image description here






acid-base







share|improve this question









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














1












1








1





$begingroup$


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.



enter image description here






acid-base







share|improve this question









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.



enter image description here






acid-base




acid-base






share|improve this question









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.











share|improve this question









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.









share|improve this question




share|improve this question








edited 1 hour ago









Karsten Theis

5,282644




5,282644






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asked 3 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.
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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














  • 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










1 Answer
1






active

oldest

votes


















3












$begingroup$

$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$



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.






share|improve this answer











$endgroup$









  • 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












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1 Answer
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1 Answer
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3












$begingroup$

$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$



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.






share|improve this answer











$endgroup$









  • 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
















3












$begingroup$

$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$



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.






share|improve this answer











$endgroup$









  • 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














3












3








3





$begingroup$

$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$



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.






share|improve this answer











$endgroup$



$$
begin{align}
ce{HNO2 + H2O &<=> NO2- + H3O+} &quad K &= K_mathrm{a} tag{1}\
ce{NO2- + H2O &<=> HNO2 + OH-} &quad K &= K_mathrm{b} tag{2}\
ce{H2O + H2O &<=> H3O+ + OH-} &quad K &= K_mathrm{w} tag{3}
end{align}
$$



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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 2 hours 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














  • 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










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