Equilibrium constant for the neutralization of weak acid by strong base
$begingroup$
Acetic acid has a $K_mathrm{a}$ of $pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $ce{NaOH}$?
Given acetic acid
$$ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} qquad K_mathrm{a} = pu{1.8e-5}$$
$$ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_mathrm{w} = K_mathrm{a}K_mathrm{b}$, then we get $K_mathrm{b} = pu{5.55e-10}$. How do I use this to find $K_mathrm{eq}$?
I know how to find $K_mathrm{eq}$ 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
17.2k655116
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
add a comment |
$begingroup$
Acetic acid has a $K_mathrm{a}$ of $pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $ce{NaOH}$?
Given acetic acid
$$ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} qquad K_mathrm{a} = pu{1.8e-5}$$
$$ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_mathrm{w} = K_mathrm{a}K_mathrm{b}$, then we get $K_mathrm{b} = pu{5.55e-10}$. How do I use this to find $K_mathrm{eq}$?
I know how to find $K_mathrm{eq}$ 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
17.2k655116
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
add a comment |
$begingroup$
Acetic acid has a $K_mathrm{a}$ of $pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $ce{NaOH}$?
Given acetic acid
$$ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} qquad K_mathrm{a} = pu{1.8e-5}$$
$$ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_mathrm{w} = K_mathrm{a}K_mathrm{b}$, then we get $K_mathrm{b} = pu{5.55e-10}$. How do I use this to find $K_mathrm{eq}$?
I know how to find $K_mathrm{eq}$ 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
17.2k655116
asked Feb 23 at 6:52
AvarosaAvarosa
62
$endgroup$
Acetic acid has a $K_mathrm{a}$ of $pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $ce{NaOH}$?
Given acetic acid
$$ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} qquad K_mathrm{a} = pu{1.8e-5}$$
$$ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_mathrm{w} = K_mathrm{a}K_mathrm{b}$, then we get $K_mathrm{b} = pu{5.55e-10}$. How do I use this to find $K_mathrm{eq}$?
I know how to find $K_mathrm{eq}$ 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
17.2k655116
asked Feb 23 at 6:52
AvarosaAvarosa
62
edited Feb 23 at 7:13
andselisk
17.2k655116
asked Feb 23 at 6:52
AvarosaAvarosa
62
edited Feb 23 at 7:13
andselisk
17.2k655116
edited Feb 23 at 7:13
andselisk
17.2k655116
edited Feb 23 at 7:13
andselisk
17.2k655116
17.2k655116
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:
$$ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$
Since $[ce{H2O}] = text{const}$ (reaction medium), $K'[ce{H2O}] = K = text{const}$:
$$K = frac{[ce{OAc-}]}{[ce{HOAc}][ce{OH-}]}$$
By multiplying both numerator and denominator by $[ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrm{a}$ and ionic product of water $K_mathrm{w}$:
$$K = frac{color{red}{[ce{OAc-}][ce{H+}]}}{color{red}{[ce{HOAc}]}[ce{OH-}][ce{H+}]} = frac{color{red}{K_mathrm{a}}}{K_mathrm{w}}$$
For acetic acid:
$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
$endgroup$
$begingroup$
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
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1 Answer
1
active
oldest
votes
active
oldest
votes
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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:
$$ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$
Since $[ce{H2O}] = text{const}$ (reaction medium), $K'[ce{H2O}] = K = text{const}$:
$$K = frac{[ce{OAc-}]}{[ce{HOAc}][ce{OH-}]}$$
By multiplying both numerator and denominator by $[ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrm{a}$ and ionic product of water $K_mathrm{w}$:
$$K = frac{color{red}{[ce{OAc-}][ce{H+}]}}{color{red}{[ce{HOAc}]}[ce{OH-}][ce{H+}]} = frac{color{red}{K_mathrm{a}}}{K_mathrm{w}}$$
For acetic acid:
$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
$endgroup$
$begingroup$
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:
$$ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$
Since $[ce{H2O}] = text{const}$ (reaction medium), $K'[ce{H2O}] = K = text{const}$:
$$K = frac{[ce{OAc-}]}{[ce{HOAc}][ce{OH-}]}$$
By multiplying both numerator and denominator by $[ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrm{a}$ and ionic product of water $K_mathrm{w}$:
$$K = frac{color{red}{[ce{OAc-}][ce{H+}]}}{color{red}{[ce{HOAc}]}[ce{OH-}][ce{H+}]} = frac{color{red}{K_mathrm{a}}}{K_mathrm{w}}$$
For acetic acid:
$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
$endgroup$
$begingroup$
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:
$$ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$
Since $[ce{H2O}] = text{const}$ (reaction medium), $K'[ce{H2O}] = K = text{const}$:
$$K = frac{[ce{OAc-}]}{[ce{HOAc}][ce{OH-}]}$$
By multiplying both numerator and denominator by $[ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrm{a}$ and ionic product of water $K_mathrm{w}$:
$$K = frac{color{red}{[ce{OAc-}][ce{H+}]}}{color{red}{[ce{HOAc}]}[ce{OH-}][ce{H+}]} = frac{color{red}{K_mathrm{a}}}{K_mathrm{w}}$$
For acetic acid:
$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
$endgroup$
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:
$$ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$
Since $[ce{H2O}] = text{const}$ (reaction medium), $K'[ce{H2O}] = K = text{const}$:
$$K = frac{[ce{OAc-}]}{[ce{HOAc}][ce{OH-}]}$$
By multiplying both numerator and denominator by $[ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_mathrm{a}$ and ionic product of water $K_mathrm{w}$:
$$K = frac{color{red}{[ce{OAc-}][ce{H+}]}}{color{red}{[ce{HOAc}]}[ce{OH-}][ce{H+}]} = frac{color{red}{K_mathrm{a}}}{K_mathrm{w}}$$
For acetic acid:
$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
edited Feb 23 at 7:15
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
answered Feb 23 at 7:09
andseliskandselisk
17.2k655116
17.2k655116
$begingroup$
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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