Equilibrium constant for the neutralization of weak acid by strong base
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Acetic acid has a $K_mathrma$ of $pu1.8e-5$. What is the equilibrium constant for the neutralization of this acid with $ceNaOH$?
Given acetic acid
$$ceHC2H3O2 + H2O <=> C2H3O2- + H3O+ qquad K_mathrma = pu1.8e-5$$
$$ceHC2H3O2 + OH- <=> C2H3O2- + H2O$$
So, if we do $K_mathrmw = K_mathrmaK_mathrmb$, then we get $K_mathrmb = pu5.55e-10$. How do I use this to find $K_mathrmeq$?
I know how to find $K_mathrmeq$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?
acid-base equilibrium aqueous-solution
edited Feb 23 at 7:13
andselisk
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
add a comment |
$begingroup$
Acetic acid has a $K_mathrma$ of $pu1.8e-5$. What is the equilibrium constant for the neutralization of this acid with $ceNaOH$?
Given acetic acid
$$ceHC2H3O2 + H2O <=> C2H3O2- + H3O+ qquad K_mathrma = pu1.8e-5$$
$$ceHC2H3O2 + OH- <=> C2H3O2- + H2O$$
So, if we do $K_mathrmw = K_mathrmaK_mathrmb$, then we get $K_mathrmb = pu5.55e-10$. How do I use this to find $K_mathrmeq$?
I know how to find $K_mathrmeq$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?
acid-base equilibrium aqueous-solution
edited Feb 23 at 7:13
andselisk
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
add a comment |
$begingroup$
Acetic acid has a $K_mathrma$ of $pu1.8e-5$. What is the equilibrium constant for the neutralization of this acid with $ceNaOH$?
Given acetic acid
$$ceHC2H3O2 + H2O <=> C2H3O2- + H3O+ qquad K_mathrma = pu1.8e-5$$
$$ceHC2H3O2 + OH- <=> C2H3O2- + H2O$$
So, if we do $K_mathrmw = K_mathrmaK_mathrmb$, then we get $K_mathrmb = pu5.55e-10$. How do I use this to find $K_mathrmeq$?
I know how to find $K_mathrmeq$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?
acid-base equilibrium aqueous-solution
edited Feb 23 at 7:13
andselisk
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
Acetic acid has a $K_mathrma$ of $pu1.8e-5$. What is the equilibrium constant for the neutralization of this acid with $ceNaOH$?
Given acetic acid
$$ceHC2H3O2 + H2O <=> C2H3O2- + H3O+ qquad K_mathrma = pu1.8e-5$$
$$ceHC2H3O2 + OH- <=> C2H3O2- + H2O$$
So, if we do $K_mathrmw = K_mathrmaK_mathrmb$, then we get $K_mathrmb = pu5.55e-10$. How do I use this to find $K_mathrmeq$?
I know how to find $K_mathrmeq$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?
acid-base equilibrium aqueous-solution
acid-base equilibrium aqueous-solution
edited Feb 23 at 7:13
andselisk
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
edited Feb 23 at 7:13
andselisk
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
edited Feb 23 at 7:13
andselisk
18k656119
edited Feb 23 at 7:13
andselisk
18k656119
edited Feb 23 at 7:13
andselisk
18k656119
18k656119
asked Feb 23 at 6:52
AvarosaAvarosa
62
asked Feb 23 at 6:52
AvarosaAvarosa
62
asked Feb 23 at 6:52
AvarosaAvarosa
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$ceHOAc + OH- <=> OAc- + H2O$$
$$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$
Since $[ceH2O] = textconst$ (reaction medium), $K'[ceH2O] = K = textconst$:
$$K = frac[ceOAc-][ceHOAc][ceOH-]$$
By multiplying both numerator and denominator by $[ceH+]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrma$ and ionic product of water $K_mathrmw$:
$$K = fraccolorred[ceOAc-][ceH+]colorred[ceHOAc][ceOH-][ceH+] = fraccolorredK_mathrmaK_mathrmw$$
For acetic acid:
$$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$
answered Feb 23 at 7:09
andseliskandselisk
18k656119
$endgroup$
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$ceHOAc + OH- <=> OAc- + H2O$$
$$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$
Since $[ceH2O] = textconst$ (reaction medium), $K'[ceH2O] = K = textconst$:
$$K = frac[ceOAc-][ceHOAc][ceOH-]$$
By multiplying both numerator and denominator by $[ceH+]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrma$ and ionic product of water $K_mathrmw$:
$$K = fraccolorred[ceOAc-][ceH+]colorred[ceHOAc][ceOH-][ceH+] = fraccolorredK_mathrmaK_mathrmw$$
For acetic acid:
$$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$
answered Feb 23 at 7:09
andseliskandselisk
18k656119
$endgroup$
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
add a comment |
$begingroup$
Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$ceHOAc + OH- <=> OAc- + H2O$$
$$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$
Since $[ceH2O] = textconst$ (reaction medium), $K'[ceH2O] = K = textconst$:
$$K = frac[ceOAc-][ceHOAc][ceOH-]$$
By multiplying both numerator and denominator by $[ceH+]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrma$ and ionic product of water $K_mathrmw$:
$$K = fraccolorred[ceOAc-][ceH+]colorred[ceHOAc][ceOH-][ceH+] = fraccolorredK_mathrmaK_mathrmw$$
For acetic acid:
$$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$
answered Feb 23 at 7:09
andseliskandselisk
18k656119
$endgroup$
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
add a comment |
$begingroup$
Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$ceHOAc + OH- <=> OAc- + H2O$$
$$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$
Since $[ceH2O] = textconst$ (reaction medium), $K'[ceH2O] = K = textconst$:
$$K = frac[ceOAc-][ceHOAc][ceOH-]$$
By multiplying both numerator and denominator by $[ceH+]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrma$ and ionic product of water $K_mathrmw$:
$$K = fraccolorred[ceOAc-][ceH+]colorred[ceHOAc][ceOH-][ceH+] = fraccolorredK_mathrmaK_mathrmw$$
For acetic acid:
$$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$
answered Feb 23 at 7:09
andseliskandselisk
18k656119
$endgroup$
Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$ceHOAc + OH- <=> OAc- + H2O$$
$$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$
Since $[ceH2O] = textconst$ (reaction medium), $K'[ceH2O] = K = textconst$:
$$K = frac[ceOAc-][ceHOAc][ceOH-]$$
By multiplying both numerator and denominator by $[ceH+]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrma$ and ionic product of water $K_mathrmw$:
$$K = fraccolorred[ceOAc-][ceH+]colorred[ceHOAc][ceOH-][ceH+] = fraccolorredK_mathrmaK_mathrmw$$
For acetic acid:
$$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$
answered Feb 23 at 7:09
andseliskandselisk
18k656119
edited Feb 23 at 7:15
answered Feb 23 at 7:09
andseliskandselisk
18k656119
answered Feb 23 at 7:09
andseliskandselisk
18k656119
answered Feb 23 at 7:09
andseliskandselisk
18k656119
18k656119
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
add a comment |
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.
$endgroup$
– Avarosa
Feb 23 at 7:54
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.
$endgroup$
– andselisk
Feb 23 at 7:58
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
$begingroup$
Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT!
$endgroup$
– Avarosa
Feb 23 at 18:56
add a comment |
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