Equilibrium constant for the neutralization of weak acid by strong base












0












$begingroup$



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











$endgroup$

















    0












    $begingroup$



    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











    $endgroup$















      0












      0








      0





      $begingroup$



      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











      $endgroup$





      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 23 at 7:13









      andselisk

      17.2k655116




      17.2k655116










      asked Feb 23 at 6:52









      AvarosaAvarosa

      62




      62






















          1 Answer
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          3












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          3












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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











          $endgroup$



          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:



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







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Feb 23 at 7:15

























          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


















          • $begingroup$
            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


















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