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
          1






          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:



          $$ce{HOAc + OH- <=> OAc- + H2O}$$



          $$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$



          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "431"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f109948%2fequilibrium-constant-for-the-neutralization-of-weak-acid-by-strong-base%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          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:



          $$ce{HOAc + OH- <=> OAc- + H2O}$$



          $$K' = frac{[ce{OAc-}][ce{H2O}]}{[ce{HOAc}][ce{OH-}]}$$



          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Chemistry Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f109948%2fequilibrium-constant-for-the-neutralization-of-weak-acid-by-strong-base%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
          Read more