Equilibrium constant for the neutralization of weak acid by strong base

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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?










share|improve this question











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    0












    $begingroup$



    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?










    share|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$



      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?










      share|improve this question











      $endgroup$





      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 23 at 7:13









      andselisk

      18k656119




      18k656119










      asked Feb 23 at 6:52









      AvarosaAvarosa

      62




      62




















          1 Answer
          1






          active

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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:



          $$ceHOAc + OH- <=> OAc- + H2O$$



          $$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$



          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:



          $$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$






          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
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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:



          $$ceHOAc + OH- <=> OAc- + H2O$$



          $$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$



          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:



          $$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$






          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:



          $$ceHOAc + OH- <=> OAc- + H2O$$



          $$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$



          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:



          $$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$






          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:



          $$ceHOAc + OH- <=> OAc- + H2O$$



          $$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$



          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:



          $$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$






          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:



          $$ceHOAc + OH- <=> OAc- + H2O$$



          $$K' = frac[ceOAc-][ceH2O][ceHOAc][ceOH-]$$



          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:



          $$K = fracpu1.8e-5pu1e-14 = pu1.8e9$$







          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

          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
















          • $begingroup$
            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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