Solubility of a tribasic weak acid












4












$begingroup$


I have a question that reads:




$ce{H3A}$ is a tribasic acid. The three deprotonations can be written as



$ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$



What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?



a) $ce{H3A}$



b) $ce{H2A-}$



c) $ce{HA^{2-}}$



d) $ce{A^{3-}}$




I worked this out as



$ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$



At pH 5, $ce{[H+] = 10^{-5}M}$



So,



$ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$



and similarly,



$ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$



$ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$



Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).



This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).






solubility







share|improve this question









New contributor




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







$endgroup$

















    4












    $begingroup$


    I have a question that reads:




    $ce{H3A}$ is a tribasic acid. The three deprotonations can be written as



    $ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$



    What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?



    a) $ce{H3A}$



    b) $ce{H2A-}$



    c) $ce{HA^{2-}}$



    d) $ce{A^{3-}}$




    I worked this out as



    $ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$



    At pH 5, $ce{[H+] = 10^{-5}M}$



    So,



    $ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$



    and similarly,



    $ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$



    $ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$



    Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).



    This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).






    solubility







    share|improve this question









    New contributor




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







    $endgroup$















      4












      4








      4





      $begingroup$


      I have a question that reads:




      $ce{H3A}$ is a tribasic acid. The three deprotonations can be written as



      $ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$



      What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?



      a) $ce{H3A}$



      b) $ce{H2A-}$



      c) $ce{HA^{2-}}$



      d) $ce{A^{3-}}$




      I worked this out as



      $ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$



      At pH 5, $ce{[H+] = 10^{-5}M}$



      So,



      $ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$



      and similarly,



      $ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$



      $ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$



      Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).



      This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).






      solubility







      share|improve this question









      New contributor




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







      $endgroup$




      I have a question that reads:




      $ce{H3A}$ is a tribasic acid. The three deprotonations can be written as



      $ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$



      What is the predominant species in a 0.10M solution of $ce{H3A}$ at pH 5?



      a) $ce{H3A}$



      b) $ce{H2A-}$



      c) $ce{HA^{2-}}$



      d) $ce{A^{3-}}$




      I worked this out as



      $ce{Ka1 = frac{[H+][H2A-]}{[H3A]}}, ce{Ka2 = frac{[H+][HA^{2-}]}{[H2A-]}}, ce{Ka3 = frac{[H+][A^{3-}]}{[HA^{2-}]}}$



      At pH 5, $ce{[H+] = 10^{-5}M}$



      So,



      $ce{[H+] = 10^{-5} = Ka1frac{[H2A-]}{[H3A]} = 10^{-2.9} frac{[H2A-]}{[H3A]}}ce{implies frac{[H2A-]}{[H3A]} = 10^{2.1} }$



      and similarly,



      $ce{frac{[HA2-]}{[H2A-]} = 10^{0.8}}$



      $ce{frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$



      Since $ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $ce{HA^{2-}}$, so c).



      This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).






      solubility




      solubility






      share|improve this question









      New contributor




      Gremlin 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









      New contributor




      Gremlin 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




      share|improve this question








      edited 12 hours ago







      Gremlin













      New contributor




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









      asked 13 hours ago









      GremlinGremlin

      1236




      1236




      New contributor




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





      New contributor





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






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






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          You can also use the Henderson-Hasselbalch equation:



          $$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$



          rearranges to:



          $$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$



          $log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.



          $log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.



          $log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.



          The equilibrium of interest is the third dissociation.



          Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.






          share|improve this answer









          $endgroup$













            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
            });


            }
            });






            Gremlin is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f110253%2fsolubility-of-a-tribasic-weak-acid%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









            5












            $begingroup$

            You can also use the Henderson-Hasselbalch equation:



            $$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$



            rearranges to:



            $$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$



            $log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.



            $log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.



            $log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.



            The equilibrium of interest is the third dissociation.



            Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.






            share|improve this answer









            $endgroup$


















              5












              $begingroup$

              You can also use the Henderson-Hasselbalch equation:



              $$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$



              rearranges to:



              $$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$



              $log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.



              $log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.



              $log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.



              The equilibrium of interest is the third dissociation.



              Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.






              share|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                You can also use the Henderson-Hasselbalch equation:



                $$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$



                rearranges to:



                $$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$



                $log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.



                $log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.



                $log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.



                The equilibrium of interest is the third dissociation.



                Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.






                share|improve this answer









                $endgroup$



                You can also use the Henderson-Hasselbalch equation:



                $$mathrm{pH} = mathrm{p}K_{mathrm{a}} + log frac{ce{[A-]}}{ce{[HA]}}$$



                rearranges to:



                $$mathrm{pH} - mathrm{p}K_{mathrm{a}} = log frac{ce{[A-]}}{ce{[HA]}}$$



                $log frac{ce{[A-]}}{ce{[HA]}} > 0$ for conjugate base dominating.



                $log frac{ce{[A-]}}{ce{[HA]}} < 0$ for acid dominating.



                $log frac{ce{[A-]}}{ce{[HA]}} = 0$ if they're present in the same concentrations.



                The equilibrium of interest is the third dissociation.



                Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($ce{HA^{2-}}$) dominates over the conjugate base ($ce{A^{3-}}$). You should be able to extend this for each equilibrium.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 13 hours ago









                ZheZhe

                12.7k12550




                12.7k12550






















                    Gremlin is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    Gremlin is a new contributor. Be nice, and check out our Code of Conduct.













                    Gremlin is a new contributor. Be nice, and check out our Code of Conduct.












                    Gremlin is a new contributor. Be nice, and check out our Code of Conduct.
















                    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%2f110253%2fsolubility-of-a-tribasic-weak-acid%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

                    "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness. ...
                    Read more

                    Alcedinidae

                    Image
                    Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
                    Read more

                    Origin of the phrase “under your belt”?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
                    Read more