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Equilibrium constant for the neutralization of weak acid by strong base

Acid dissociation constant and equilibrium constantHow to calculate the equilibrium constant for an esterification?Neutralization reaction with insoluble salt (weak acid, strong base)?Why is water not part of the equilibrium constant?When do I use the hydronium ion in an acid-base equation?Calculating the pH of the endpoint in a titration of weak acid and strong baseWhy is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?Acid-base titration: Calculate pKa with only three values givenCalculating PH of solution after adding strong acid (equilibrium)How to calculate the equilibrium constant of acetic acid-hydroxide ion equilibrium?

1

$begingroup$

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

share|improve this question

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

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

share|improve this question

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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

share|improve this question

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa 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

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

andselisk

16.5k654115

edited 2 hours ago

andselisk

16.5k654115

asked 3 hours ago

AvarosaAvarosa

61

New contributor

Avarosa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

AvarosaAvarosa

61

1 Answer
1

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2

$begingroup$

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:

$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$

share|improve this answer

edited 2 hours ago

answered 3 hours ago

andseliskandselisk

16.5k654115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

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:

$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$

share|improve this answer

edited 2 hours ago

answered 3 hours ago

andseliskandselisk

16.5k654115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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:

$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$

share|improve this answer

edited 2 hours ago

answered 3 hours ago

andseliskandselisk

16.5k654115

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  
  
  

  
  
  
  

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:

$$K = frac{pu{1.8e-5}}{pu{1e-14}} = pu{1.8e9}$$

share|improve this answer

edited 2 hours ago

answered 3 hours ago

andseliskandselisk

16.5k654115

$endgroup$

share|improve this answer

edited 2 hours ago

answered 3 hours ago

andseliskandselisk

16.5k654115

answered 3 hours ago

andseliskandselisk

16.5k654115

answered 3 hours ago

andseliskandselisk

16.5k654115