When to use 1/Ka vs Kb [on hold]How does one calculate the pH of an aqueous solution of a weak acid and a weak base?Comparing acidities of substituted and aromatic carboxylic acidsWhat is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?How do I write out the chemical reaction of ammonia with strong acid?Apparent solubility of Ag2C2O4 in a buffer solutionIdentifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values?pH of 0.5M K2CO3How to calculate the pH of the neutralisation of HCN with excess of KOH?Acid-base titration: Calculate pKa with only three values givenHow accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt?
Can an isometry leave entropy invariant?
I drew a randomly colored grid of points with tikz, how do I force it to remember the first grid from then on?
What are the differences between credential stuffing and password spraying?
Why isn't nylon as strong as kevlar?
How to model the curly cable part of the phone
On which topic did Indiana Jones write his doctoral thesis?
Can my company stop me from working overtime?
Prove that the limit exists or does not exist
Why was the battle set up *outside* Winterfell?
How can modem speed be 10 times slower than router?
Verb "geeitet" in an old scientific text
Which module had more 'comfort' in terms of living space, the Lunar Module or the Command module?
Why Isn’t SQL More Refactorable?
What is the difference between 反日 and 日本たたき?
What is the name of this hexagon/pentagon polyhedron?
Expressing 'our' for objects belonging to our apartment
Why didn't the check-in agent recognize my long term visa?
What matters more when it comes to book covers? Is it ‘professional quality’ or relevancy?
Does a card have a keyword if it has the same effect as said keyword?
How do LIGO and VIRGO know that a gravitational wave has its origin in a neutron star or a black hole?
Why is B♯ higher than C♭ in 31-ET?
How I can I roll a number of non-digital dice to get a random number between 1 and 150?
Can you complete the sequence?
How can I support myself financially as a 17 year old with a loan?
When to use 1/Ka vs Kb [on hold]
How does one calculate the pH of an aqueous solution of a weak acid and a weak base?Comparing acidities of substituted and aromatic carboxylic acidsWhat is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?How do I write out the chemical reaction of ammonia with strong acid?Apparent solubility of Ag2C2O4 in a buffer solutionIdentifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values?pH of 0.5M K2CO3How to calculate the pH of the neutralisation of HCN with excess of KOH?Acid-base titration: Calculate pKa with only three values givenHow accurately does (pKa1 + pKa2) / 2 estimate the pH of an amphoteric salt?
$begingroup$
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 $K_mathrma$ and $K_mathrmb$. For example here is a calculation where for the reaction involving $ceNO2-$, the equilibrium constant is $1/K_mathrma$ even though it shows $ceNO2-$ acting as a base.
Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)
$ceNO2-(aq)$, $ceF-(aq)$, $ceHF(aq)$, $ceHNO2(aq)$
$ceHNO2$ $(K_mathrma = pu4.6e-4)$ is a weaker acid than $ceHF$ $(K_mathrma = pu6.6e-4)$, so $ceHNO2$ holds onto its proton more tightly than
$ceHF$. So if we write a reaction when $ceNO2-$ takes a proton from $ceF-$, we'll have a reaction with $K > 1$:
$$ceHF + NO2- → HNO2 + F-$$
We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:
$$
beginalign
ceHF &→ H+ + F- &quad &K_mathrma(ceHF)\
ceH+ + NO2- &→ HNO2 &quad &1/K_mathrma(ceHNO2)\
hline
ceHF + NO2- &→ HNO2 + F- &quad &fracK_mathrma(ceHF)K_mathrma(ceHNO2) = fracpu6.6e-4pu4.6e-4 = 1.4 > 1
endalign
$$
acid-base equilibrium phy
edited 2 days ago
andselisk
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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.
$endgroup$
put on hold as unclear what you're asking by Todd Minehardt, Mithoron, airhuff, Jon Custer, aventurin 13 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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 $K_mathrma$ and $K_mathrmb$. For example here is a calculation where for the reaction involving $ceNO2-$, the equilibrium constant is $1/K_mathrma$ even though it shows $ceNO2-$ acting as a base.
Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)
$ceNO2-(aq)$, $ceF-(aq)$, $ceHF(aq)$, $ceHNO2(aq)$
$ceHNO2$ $(K_mathrma = pu4.6e-4)$ is a weaker acid than $ceHF$ $(K_mathrma = pu6.6e-4)$, so $ceHNO2$ holds onto its proton more tightly than
$ceHF$. So if we write a reaction when $ceNO2-$ takes a proton from $ceF-$, we'll have a reaction with $K > 1$:
$$ceHF + NO2- → HNO2 + F-$$
We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:
$$
beginalign
ceHF &→ H+ + F- &quad &K_mathrma(ceHF)\
ceH+ + NO2- &→ HNO2 &quad &1/K_mathrma(ceHNO2)\
hline
ceHF + NO2- &→ HNO2 + F- &quad &fracK_mathrma(ceHF)K_mathrma(ceHNO2) = fracpu6.6e-4pu4.6e-4 = 1.4 > 1
endalign
$$
acid-base equilibrium phy
edited 2 days ago
andselisk
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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.
$endgroup$
put on hold as unclear what you're asking by Todd Minehardt, Mithoron, airhuff, Jon Custer, aventurin 13 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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 $K_mathrma$ and $K_mathrmb$. For example here is a calculation where for the reaction involving $ceNO2-$, the equilibrium constant is $1/K_mathrma$ even though it shows $ceNO2-$ acting as a base.
Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)
$ceNO2-(aq)$, $ceF-(aq)$, $ceHF(aq)$, $ceHNO2(aq)$
$ceHNO2$ $(K_mathrma = pu4.6e-4)$ is a weaker acid than $ceHF$ $(K_mathrma = pu6.6e-4)$, so $ceHNO2$ holds onto its proton more tightly than
$ceHF$. So if we write a reaction when $ceNO2-$ takes a proton from $ceF-$, we'll have a reaction with $K > 1$:
$$ceHF + NO2- → HNO2 + F-$$
We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:
$$
beginalign
ceHF &→ H+ + F- &quad &K_mathrma(ceHF)\
ceH+ + NO2- &→ HNO2 &quad &1/K_mathrma(ceHNO2)\
hline
ceHF + NO2- &→ HNO2 + F- &quad &fracK_mathrma(ceHF)K_mathrma(ceHNO2) = fracpu6.6e-4pu4.6e-4 = 1.4 > 1
endalign
$$
acid-base equilibrium phy
edited 2 days ago
andselisk
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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 $K_mathrma$ and $K_mathrmb$. For example here is a calculation where for the reaction involving $ceNO2-$, the equilibrium constant is $1/K_mathrma$ even though it shows $ceNO2-$ acting as a base.
Use the following four aqueous species to write an acid-base reaction with $K > 1$. (An acid-base reaction is one that involves the exchange of a proton. Your reaction will involve all four species.)
$ceNO2-(aq)$, $ceF-(aq)$, $ceHF(aq)$, $ceHNO2(aq)$
$ceHNO2$ $(K_mathrma = pu4.6e-4)$ is a weaker acid than $ceHF$ $(K_mathrma = pu6.6e-4)$, so $ceHNO2$ holds onto its proton more tightly than
$ceHF$. So if we write a reaction when $ceNO2-$ takes a proton from $ceF-$, we'll have a reaction with $K > 1$:
$$ceHF + NO2- → HNO2 + F-$$
We can get the $K$ for this reaction that adding the two acid dissocation reactions together as:
$$
beginalign
ceHF &→ H+ + F- &quad &K_mathrma(ceHF)\
ceH+ + NO2- &→ HNO2 &quad &1/K_mathrma(ceHNO2)\
hline
ceHF + NO2- &→ HNO2 + F- &quad &fracK_mathrma(ceHF)K_mathrma(ceHNO2) = fracpu6.6e-4pu4.6e-4 = 1.4 > 1
endalign
$$
acid-base equilibrium phy
acid-base equilibrium phy
edited 2 days ago
andselisk
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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 2 days ago
andselisk
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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 2 days ago
andselisk
20.4k669132
edited 2 days ago
andselisk
20.4k669132
edited 2 days ago
andselisk
20.4k669132
20.4k669132
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
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 Apr 29 at 1:38
Lucky LucyLucky Lucy
211
asked Apr 29 at 1:38
Lucky LucyLucky Lucy
211
211
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.
put on hold as unclear what you're asking by Todd Minehardt, Mithoron, airhuff, Jon Custer, aventurin 13 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Todd Minehardt, Mithoron, airhuff, Jon Custer, aventurin 13 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$
beginalign
ceHNO2 + H2O &<=> NO2- + H3O+ &quad K &= K_mathrma tag1\
ceNO2- + H2O &<=> HNO2 + OH- &quad K &= K_mathrmb tag2\
ceH2O + H2O &<=> H3O+ + OH- &quad K &= K_mathrmw tag3
endalign
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrma$, $K_mathrmb$, and $K_mathrmw$. If you reverse the first reaction, you get:
$$ceNO2- + H3O+ <=> HNO2 + H2O qquad K = 1/K_mathrmatag1a$$
(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:
$$K_mathrma times K_mathrmb = K_mathrmw$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrma$ vs $K_mathrmb$
Use the equilibrium constant that matches the reaction you are working on.
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
beginalign
ceHNO2 + H2O &<=> NO2- + H3O+ &quad K &= K_mathrma tag1\
ceNO2- + H2O &<=> HNO2 + OH- &quad K &= K_mathrmb tag2\
ceH2O + H2O &<=> H3O+ + OH- &quad K &= K_mathrmw tag3
endalign
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrma$, $K_mathrmb$, and $K_mathrmw$. If you reverse the first reaction, you get:
$$ceNO2- + H3O+ <=> HNO2 + H2O qquad K = 1/K_mathrmatag1a$$
(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:
$$K_mathrma times K_mathrmb = K_mathrmw$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrma$ vs $K_mathrmb$
Use the equilibrium constant that matches the reaction you are working on.
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
$endgroup$
add a comment |
$begingroup$
$$
beginalign
ceHNO2 + H2O &<=> NO2- + H3O+ &quad K &= K_mathrma tag1\
ceNO2- + H2O &<=> HNO2 + OH- &quad K &= K_mathrmb tag2\
ceH2O + H2O &<=> H3O+ + OH- &quad K &= K_mathrmw tag3
endalign
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrma$, $K_mathrmb$, and $K_mathrmw$. If you reverse the first reaction, you get:
$$ceNO2- + H3O+ <=> HNO2 + H2O qquad K = 1/K_mathrmatag1a$$
(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:
$$K_mathrma times K_mathrmb = K_mathrmw$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrma$ vs $K_mathrmb$
Use the equilibrium constant that matches the reaction you are working on.
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
$endgroup$
add a comment |
$begingroup$
$$
beginalign
ceHNO2 + H2O &<=> NO2- + H3O+ &quad K &= K_mathrma tag1\
ceNO2- + H2O &<=> HNO2 + OH- &quad K &= K_mathrmb tag2\
ceH2O + H2O &<=> H3O+ + OH- &quad K &= K_mathrmw tag3
endalign
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrma$, $K_mathrmb$, and $K_mathrmw$. If you reverse the first reaction, you get:
$$ceNO2- + H3O+ <=> HNO2 + H2O qquad K = 1/K_mathrmatag1a$$
(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:
$$K_mathrma times K_mathrmb = K_mathrmw$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrma$ vs $K_mathrmb$
Use the equilibrium constant that matches the reaction you are working on.
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
$endgroup$
$$
beginalign
ceHNO2 + H2O &<=> NO2- + H3O+ &quad K &= K_mathrma tag1\
ceNO2- + H2O &<=> HNO2 + OH- &quad K &= K_mathrmb tag2\
ceH2O + H2O &<=> H3O+ + OH- &quad K &= K_mathrmw tag3
endalign
$$
Above are the reactions associated with equilibrium constants commonly called $K_mathrma$, $K_mathrmb$, and $K_mathrmw$. If you reverse the first reaction, you get:
$$ceNO2- + H3O+ <=> HNO2 + H2O qquad K = 1/K_mathrmatag1a$$
(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:
$$K_mathrma times K_mathrmb = K_mathrmw$$
This is because when adding up the first and the second reaction, you get the third.
When to use $1/K_mathrma$ vs $K_mathrmb$
Use the equilibrium constant that matches the reaction you are working on.
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
edited Apr 29 at 3:18
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
answered Apr 29 at 2:44
Karsten TheisKarsten Theis
5,595745
5,595745
add a comment |
add a comment |
