{"id":2536,"date":"2018-06-19T20:34:54","date_gmt":"2018-06-19T20:34:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/1-8-2-bronsted-lowry-acids-and-bases\/"},"modified":"2018-08-06T12:44:45","modified_gmt":"2018-08-06T12:44:45","slug":"6-3-bronsted-lowry-acids-bases","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-3-bronsted-lowry-acids-bases\/","title":{"raw":"6.3. Bronsted-Lowry acids\/bases","rendered":"6.3. Bronsted-Lowry acids\/bases"},"content":{"raw":"<section class=\"mt-content-container\">\r\n<div id=\"s1281\" class=\"mt-include\">\r\n<div class=\"mt-section\">\r\n<h1 class=\"editable\">A: The Br\u00f8nsted-Lowry definition of acidity and basicity<\/h1>\r\nWe\u2019ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these was proposed in 1923 by the Danish chemist Johannes Br\u00f8nsted and the English chemist Thomas Lowry, and has come to be known as the <strong>Br\u00f8nsted-Lowry definition of acidity and basicity<\/strong>.\u00a0 An acid, by the Br\u00f8nsted-Lowry definition, is a species which acts as a proton donor (i.e., it gives away an H<sup>+<\/sup>), while a base is a proton (H<sup>+<\/sup>) acceptor.\u00a0 One of the most familiar examples of a Br\u00f8nsted-Lowry acid-base reaction is between hydrochloric acid and hydroxide ion:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203455\/image001.png\" alt=\"\" width=\"385\" height=\"58\" \/>\r\n\r\nIn this reaction, a proton is transferred from HCl (the acid, or proton <em>donor<\/em>) to hydroxide ion (the base, or proton <em>acceptor<\/em>). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process.\r\n\r\nAfter a Br\u00f8nsted-Lowry acid donates a proton, what remains is called the <strong>conjugate base<\/strong>.\u00a0 Chloride ion is thus the conjugate base of hydrochloric acid. Conversely, when a Br\u00f8nsted-Lowry base accepts a proton it is converted into its <strong>conjugate acid<\/strong> form: water is thus the conjugate acid of hydroxide ion.\r\n\r\nHere is an organic acid-base reaction between acetic acid and methylamine:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203458\/image002.png\" alt=\"image002.png\" width=\"523\" height=\"123\" \/>\r\n\r\nIn the reverse of this reaction, acetate ion is the base and methylammonium ion (protonated methylamine) is the acid.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203501\/image003.png\" alt=\"\" width=\"524\" height=\"128\" \/>\r\n\r\nWhat makes a compound acidic (likely to donate a proton) or basic (likely to accept a proton)? Answering that question is one of our main jobs in this chapter, and will require us to put to use much of what we learned about organic structure in the first two chapters, as well as the ideas about thermodynamics that we reviewed in chapter 5.\r\n\r\nFor now, let's just consider one common property of bases: <em>in order to act as a base, a molecule must have a reactive pair of electrons<\/em>.\u00a0 In all of the acid-base reactions we'll see in this chapter, the basic species has an atom with a lone pair of electrons. When methylamine acts as a base, for example, the lone pair of electrons on the nitrogen atom is used to form a new bond to a proton.\u00a0 A negative charge often (but not always!) indicates that a structure (in this case, an anion) is likely to act as a base.\r\n\r\nClearly, methylammonium ion cannot act as a base \u2013 it does not have a reactive pair of electrons with which to accept a proton.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203504\/image004.png\" alt=\"\" width=\"409\" height=\"123\" \/>\r\n\r\nLater, in chapter 10, we will study reactions in which a pair of electrons in a pi bond of an alkene acts in a basic fashion - but for now, will concentrate on the basicity of non-bonding (lone pair) electrons.\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\nComplete the reactions below - in other words, draw structures for the missing conjugate acids and conjugate bases that result from the curved arrows provided.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203507\/image005.png\" alt=\"\" width=\"418\" height=\"295\" \/>\r\n\r\n[reveal-answer q=\"191829\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"191829\"]<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61361\/imageE1.png?revision=1\" alt=\"\" width=\"392\" height=\"275\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"mt-section\">\u00a0In summary,<\/div>\r\n<div class=\"mt-section\">\r\n<ul>\r\n \t<li>A Br\u00f8nsted-Lowry acid is a proton (hydrogen ion) donor.<\/li>\r\n \t<li>A Br\u00f8nsted-Lowry base is a proton (hydrogen ion) acceptor.<\/li>\r\n<\/ul>\r\nWhen a Br\u00f8nsted acid HA dissociates in water, it increases the concentration of hydrogen ions in the solution, $$[H^+]$$; conversely, Br\u00f8nsted bases dissociate by taking a proton from the solvent (water) to generate $$[OH^-]$$.\r\n<ul>\r\n \t<li><em>Acid dissociation<\/em><\/li>\r\n<\/ul>\r\n\\[HA_{(aq)} \\rightleftharpoons A^-_{(aq)} + H^+_{(aq)}\\]\r\n<ul>\r\n \t<li><em>Acid Ionization Constant:<\/em><\/li>\r\n<\/ul>\r\n\\[K_a=\\dfrac{[A^-][H^+]}{[HA]}\\]\r\n<ul>\r\n \t<li><em>Base dissociation:<\/em><\/li>\r\n<\/ul>\r\n\\[B_{(aq)} + H_2O_{(l)} \\rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}\\]\r\n<ul>\r\n \t<li>Base Ionization Constant<\/li>\r\n<\/ul>\r\n\\[K_b\u00a0= \\dfrac{[HB^+][OH^-]}{[B]}\\]\r\n<p class=\"paragraph\">The determination of a substance as a Br\u00f8nsted-Lowry acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.<\/p>\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n\r\nWater does not need to be involved in a Bronsted-Lowry reaction.\u00a0 In general, for an acid HA and a base Z, we have\r\n\r\n\\[ HA + Z \\rightleftharpoons A^- + HZ^+ \\]\r\n<ul>\r\n \t<li class=\"paragraph\">A Donates H to form HZ<sup class=\"superscript\">+<\/sup>.<\/li>\r\n \t<li class=\"paragraph\">Z Accepts H from A which forms HZ<sup class=\"superscript\">+<\/sup><\/li>\r\n \t<li class=\"paragraph\">A<sup class=\"superscript\">-<\/sup> becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium<\/li>\r\n \t<li class=\"paragraph\">HZ<sup class=\"superscript\">+<\/sup> becomes a conjugate acid of Z and in the reverse reaction it donates a H to A<sup class=\"superscript\">-<\/sup> recreating Z in order to remain in equilibrium<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<div class=\"mt-section\">\r\n<div class=\"textbox exercises\">\r\n<h3>Questions<\/h3>\r\n<div class=\"mt-section\">\r\n<ol>\r\n \t<li class=\"paragraph\">Why is $$HA$$ an Acid?<\/li>\r\n \t<li class=\"paragraph\">Why is $$Z^-$$ a Base?<\/li>\r\n \t<li class=\"paragraph\">How can A<sup class=\"superscript\">-<\/sup> be a base when HA was and Acid?<\/li>\r\n \t<li class=\"paragraph\">How can HZ<sup class=\"superscript\">+<\/sup> be an acid when Z used to be a Base?<\/li>\r\n \t<li class=\"paragraph\"><strong class=\"bold\">Now that we understand the concept, let's look at an an example with actual compounds!<\/strong> \\[ HCl + H_2O \\rightleftharpoons H_3O^+ + Cl^\u00af \\]<\/li>\r\n<\/ol>\r\n<ul>\r\n \t<li class=\"paragraph\">HCl is the acid because it is donating a proton to H<sub>2<\/sub>O<\/li>\r\n \t<li class=\"paragraph\">H<sub>2<\/sub>O is the base because H<sub>2<\/sub>O is accepting a proton from HCL<\/li>\r\n \t<li class=\"paragraph\">H<sub>3<\/sub>O<sup class=\"superscript\">+<\/sup> is the conjugate acid because it is donating an acid to CL turn into it's conjugate acid H<sub>2<\/sub>O<\/li>\r\n \t<li class=\"paragraph\">Cl\u00af is the conjugate base because it accepts an H from H<sub>3<\/sub>O to return to it's conjugate acid HCl<\/li>\r\n<\/ul>\r\n<p class=\"paragraph\">How can H<sub>2<\/sub>O be a base? I thought it was neutral?<\/p>\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 id=\"Answers-1281\">Answers<\/h3>\r\n[reveal-answer q=\"454720\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"454720\"]\r\n<ol>\r\n \t<li class=\"paragraph\"><em class=\"italic\">It has a proton that can be transferred<\/em><\/li>\r\n \t<li class=\"paragraph\"><em class=\"italic\">It receives a proton from HA<\/em><\/li>\r\n \t<li class=\"paragraph\"><em class=\"italic\">A<sup class=\"superscript\">-<\/sup> is a conjugate base because it is in need of a H in order to remain in equilibrium and return to HA<\/em><\/li>\r\n \t<li class=\"paragraph\"><em class=\"italic\">HZ<sup class=\"superscript\">+<\/sup> is a conjugate acid because it needs to donate or give away its proton in order to return to it's previous state of Z<\/em><\/li>\r\n \t<li class=\"paragraph\"><em class=\"italic\">In the <\/em>Br\u00f8nsted<em class=\"italic\">-Lowry<\/em><em class=\"italic\"> Theory what makes a compound an element or a base is whether or not it donates or accepts protons. If the H<sub>2<\/sub>O was in a different problem and was instead donating an H rather than accepting an H it would be an acid!<\/em><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<div class=\"mt-section\">\r\n<h1 id=\"title\">Overview of acid-base reactions<\/h1>\r\n<img class=\"alignright size-thumbnail wp-image-4632\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01150925\/static_qr_code_without_logo1-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/>\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=jIL333CKE9A[\/embed]\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h1 class=\"editable\">An acid-base (proton transfer) reaction<\/h1>\r\nFor our first example of chemical reactivity, let\u2019s look at a very simple reaction that occurs between hydroxide ion and hydrochloric acid:\r\n<p style=\"text-align: center\">$$HCl+OH^- \\Rightarrow H_20+Cl^-$$<\/p>\r\nThis is an acid-base reaction: a proton is transferred from HCl, the acid, to hydroxide, the base. The product is water (the conjugate acid of hydroxide)\u00a0 and chloride ion (the conjugate base of HCl). You have undoubtedly seen this reaction before in general chemistry.\u00a0 Despite its simplicity (and despite the fact that the reactants and products are\u00a0<em>inorganic<\/em>\u00a0rather than organic), this reaction allows us to consider for the first time many of the fundamental ideas of organic chemistry that we will be exploring in various contexts throughout this text.\r\n\r\nKey to understanding just about any reaction mechanism is the concept of\u00a0<strong>electron density<\/strong>, and how it is connected to the\u00a0<strong>electron movement<\/strong>\u00a0(bond-breaking and bond-forming) that occurs in a reaction. The hydroxide ion \u2013 specifically, the electronegative oxygen atom in the hydroxide ion \u2013 has high electron density due to negative charge and the polarity of the hydrogen-oxygen bond.\u00a0\u00a0 The hydroxide oxygen is\u00a0<strong>electron-rich<\/strong>.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5293\/image001.png?revision=1\" alt=\"image002.png\" width=\"187\" height=\"74\" \/>\r\n\r\nThe hydrogen atom in HCl, on the other hand, has low electron density: it is\u00a0<strong>electron-poor<\/strong>.\u00a0 As you might expect, something that is electron-rich is attracted to something that is electron-poor.\u00a0 As hydroxide and HCl move closer to each other, a lone pair of electrons on the electron-rich hydroxide oxygen is attracted by the electron-poor proton of HCl, and electron movement occurs towards the proton.\u00a0 The two electrons in the hydrogen-chlorine sigma bond are repelled by this approaching hydroxide electron density, and therefore move even farther away from the proton and towards the chlorine nucleus.\u00a0 The consequence of all of this electron movement is that the hydrogen-chlorine bond is broken, as the two electrons from that bond completely break free from the 1s orbital of the hydrogen and become a lone pair in the 3p orbital of a chloride anion.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5295\/image003.png?revision=1\" alt=\"image004.png\" width=\"369\" height=\"50\" \/>\r\n\r\nAt the same time that the hydrogen-chlorine bond is breaking, a new sigma bond forms between hydrogen and oxygen, containing the two electrons that previously were a lone pair on hydroxide.\u00a0 The result of this bond formation is, of course, a water molecule.\r\n\r\nPreviously (<a class=\"mt-disabled\" title=\"section 2.2\" href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-1-electron-flow\/\" rel=\"broken\">section 6.1.<\/a>), we saw how curved arrows were used to depict electron movement that occurs in chemical reactions, where bonds are broken and new bonds are formed.\u00a0 The HCl + OH<sup>-<\/sup>\u00a0reaction, for example, is depicted by drawing two curved arrows.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5297\/image005.png?revision=1\" alt=\"image006.png\" width=\"321\" height=\"94\" \/>\r\n\r\nThe first arrow originates at one of the lone pairs on the hydroxide oxygen and points to the \u2018H\u2019 symbol in the hydrogen bromide molecule, illustrating the \u2018attack\u2019 of the oxygen lone pair and subsequent formation of the new hydrogen-oxygen bond.\u00a0 The second curved arrow originates at the hydrogen-bromine bond and points to the \u2018Br\u2019 symbol, indicating that this bond is breaking - the two electrons are \u2018leaving\u2019 and becoming a lone pair on bromide ion.\r\n\r\nIt is very important to emphasize at this point that these curved,\u00a0<em>two-barbed <\/em>arrows always represent the movement of\u00a0<em>two<\/em>\u00a0electrons.\u00a0 Most of this book will be devoted to the description of reaction mechanisms involving two-electron movement, so these full-headed arrows will become very familiar.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5299\/image007.png?revision=1\" alt=\"image008.png\" width=\"515\" height=\"95\" \/>\r\n\r\nIn the second semester, however, we will look at radical reaction mechanisms, where single-electron movement occurs.\u00a0 For these processes, a curved, single-barbed ('<strong>fish-hook<\/strong>') arrow will be used.\r\n<div>\r\n<div id=\"exercise\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\nDraw electron movement arrows to illustrate the acid-base reaction between acetic acid, CH<sub>3<\/sub>COOH, and ammonia, NH<sub>3<\/sub>.\u00a0 Draw out the full Lewis structures of reactants and products.\r\n\r\n[reveal-answer q=\"893749\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"893749\"]<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6520\/image313.png?revision=1\" alt=\"image312.png\" width=\"490\" height=\"93\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"mt-content-container\">\n<div id=\"s1281\" class=\"mt-include\">\n<div class=\"mt-section\">\n<h1 class=\"editable\">A: The Br\u00f8nsted-Lowry definition of acidity and basicity<\/h1>\n<p>We\u2019ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these was proposed in 1923 by the Danish chemist Johannes Br\u00f8nsted and the English chemist Thomas Lowry, and has come to be known as the <strong>Br\u00f8nsted-Lowry definition of acidity and basicity<\/strong>.\u00a0 An acid, by the Br\u00f8nsted-Lowry definition, is a species which acts as a proton donor (i.e., it gives away an H<sup>+<\/sup>), while a base is a proton (H<sup>+<\/sup>) acceptor.\u00a0 One of the most familiar examples of a Br\u00f8nsted-Lowry acid-base reaction is between hydrochloric acid and hydroxide ion:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203455\/image001.png\" alt=\"\" width=\"385\" height=\"58\" \/><\/p>\n<p>In this reaction, a proton is transferred from HCl (the acid, or proton <em>donor<\/em>) to hydroxide ion (the base, or proton <em>acceptor<\/em>). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process.<\/p>\n<p>After a Br\u00f8nsted-Lowry acid donates a proton, what remains is called the <strong>conjugate base<\/strong>.\u00a0 Chloride ion is thus the conjugate base of hydrochloric acid. Conversely, when a Br\u00f8nsted-Lowry base accepts a proton it is converted into its <strong>conjugate acid<\/strong> form: water is thus the conjugate acid of hydroxide ion.<\/p>\n<p>Here is an organic acid-base reaction between acetic acid and methylamine:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203458\/image002.png\" alt=\"image002.png\" width=\"523\" height=\"123\" \/><\/p>\n<p>In the reverse of this reaction, acetate ion is the base and methylammonium ion (protonated methylamine) is the acid.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203501\/image003.png\" alt=\"\" width=\"524\" height=\"128\" \/><\/p>\n<p>What makes a compound acidic (likely to donate a proton) or basic (likely to accept a proton)? Answering that question is one of our main jobs in this chapter, and will require us to put to use much of what we learned about organic structure in the first two chapters, as well as the ideas about thermodynamics that we reviewed in chapter 5.<\/p>\n<p>For now, let&#8217;s just consider one common property of bases: <em>in order to act as a base, a molecule must have a reactive pair of electrons<\/em>.\u00a0 In all of the acid-base reactions we&#8217;ll see in this chapter, the basic species has an atom with a lone pair of electrons. When methylamine acts as a base, for example, the lone pair of electrons on the nitrogen atom is used to form a new bond to a proton.\u00a0 A negative charge often (but not always!) indicates that a structure (in this case, an anion) is likely to act as a base.<\/p>\n<p>Clearly, methylammonium ion cannot act as a base \u2013 it does not have a reactive pair of electrons with which to accept a proton.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203504\/image004.png\" alt=\"\" width=\"409\" height=\"123\" \/><\/p>\n<p>Later, in chapter 10, we will study reactions in which a pair of electrons in a pi bond of an alkene acts in a basic fashion &#8211; but for now, will concentrate on the basicity of non-bonding (lone pair) electrons.<\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<p>Complete the reactions below &#8211; in other words, draw structures for the missing conjugate acids and conjugate bases that result from the curved arrows provided.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203507\/image005.png\" alt=\"\" width=\"418\" height=\"295\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q191829\">Show Solution<\/span><\/p>\n<div id=\"q191829\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61361\/imageE1.png?revision=1\" alt=\"\" width=\"392\" height=\"275\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"mt-section\">\u00a0In summary,<\/div>\n<div class=\"mt-section\">\n<ul>\n<li>A Br\u00f8nsted-Lowry acid is a proton (hydrogen ion) donor.<\/li>\n<li>A Br\u00f8nsted-Lowry base is a proton (hydrogen ion) acceptor.<\/li>\n<\/ul>\n<p>When a Br\u00f8nsted acid HA dissociates in water, it increases the concentration of hydrogen ions in the solution, $$[H^+]$$; conversely, Br\u00f8nsted bases dissociate by taking a proton from the solvent (water) to generate $$[OH^-]$$.<\/p>\n<ul>\n<li><em>Acid dissociation<\/em><\/li>\n<\/ul>\n<p>\\[HA_{(aq)} \\rightleftharpoons A^-_{(aq)} + H^+_{(aq)}\\]<\/p>\n<ul>\n<li><em>Acid Ionization Constant:<\/em><\/li>\n<\/ul>\n<p>\\[K_a=\\dfrac{[A^-][H^+]}{[HA]}\\]<\/p>\n<ul>\n<li><em>Base dissociation:<\/em><\/li>\n<\/ul>\n<p>\\[B_{(aq)} + H_2O_{(l)} \\rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}\\]<\/p>\n<ul>\n<li>Base Ionization Constant<\/li>\n<\/ul>\n<p>\\[K_b\u00a0= \\dfrac{[HB^+][OH^-]}{[B]}\\]<\/p>\n<p class=\"paragraph\">The determination of a substance as a Br\u00f8nsted-Lowry acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<p>Water does not need to be involved in a Bronsted-Lowry reaction.\u00a0 In general, for an acid HA and a base Z, we have<\/p>\n<p>\\[ HA + Z \\rightleftharpoons A^- + HZ^+ \\]<\/p>\n<ul>\n<li class=\"paragraph\">A Donates H to form HZ<sup class=\"superscript\">+<\/sup>.<\/li>\n<li class=\"paragraph\">Z Accepts H from A which forms HZ<sup class=\"superscript\">+<\/sup><\/li>\n<li class=\"paragraph\">A<sup class=\"superscript\">&#8211;<\/sup> becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium<\/li>\n<li class=\"paragraph\">HZ<sup class=\"superscript\">+<\/sup> becomes a conjugate acid of Z and in the reverse reaction it donates a H to A<sup class=\"superscript\">&#8211;<\/sup> recreating Z in order to remain in equilibrium<\/li>\n<\/ul>\n<\/div>\n<div class=\"mt-section\">\n<div class=\"mt-section\">\n<div class=\"textbox exercises\">\n<h3>Questions<\/h3>\n<div class=\"mt-section\">\n<ol>\n<li class=\"paragraph\">Why is $$HA$$ an Acid?<\/li>\n<li class=\"paragraph\">Why is $$Z^-$$ a Base?<\/li>\n<li class=\"paragraph\">How can A<sup class=\"superscript\">&#8211;<\/sup> be a base when HA was and Acid?<\/li>\n<li class=\"paragraph\">How can HZ<sup class=\"superscript\">+<\/sup> be an acid when Z used to be a Base?<\/li>\n<li class=\"paragraph\"><strong class=\"bold\">Now that we understand the concept, let&#8217;s look at an an example with actual compounds!<\/strong> \\[ HCl + H_2O \\rightleftharpoons H_3O^+ + Cl^\u00af \\]<\/li>\n<\/ol>\n<ul>\n<li class=\"paragraph\">HCl is the acid because it is donating a proton to H<sub>2<\/sub>O<\/li>\n<li class=\"paragraph\">H<sub>2<\/sub>O is the base because H<sub>2<\/sub>O is accepting a proton from HCL<\/li>\n<li class=\"paragraph\">H<sub>3<\/sub>O<sup class=\"superscript\">+<\/sup> is the conjugate acid because it is donating an acid to CL turn into it&#8217;s conjugate acid H<sub>2<\/sub>O<\/li>\n<li class=\"paragraph\">Cl\u00af is the conjugate base because it accepts an H from H<sub>3<\/sub>O to return to it&#8217;s conjugate acid HCl<\/li>\n<\/ul>\n<p class=\"paragraph\">How can H<sub>2<\/sub>O be a base? I thought it was neutral?<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 id=\"Answers-1281\">Answers<\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q454720\">Show Solution<\/span><\/p>\n<div id=\"q454720\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li class=\"paragraph\"><em class=\"italic\">It has a proton that can be transferred<\/em><\/li>\n<li class=\"paragraph\"><em class=\"italic\">It receives a proton from HA<\/em><\/li>\n<li class=\"paragraph\"><em class=\"italic\">A<sup class=\"superscript\">&#8211;<\/sup> is a conjugate base because it is in need of a H in order to remain in equilibrium and return to HA<\/em><\/li>\n<li class=\"paragraph\"><em class=\"italic\">HZ<sup class=\"superscript\">+<\/sup> is a conjugate acid because it needs to donate or give away its proton in order to return to it&#8217;s previous state of Z<\/em><\/li>\n<li class=\"paragraph\"><em class=\"italic\">In the <\/em>Br\u00f8nsted<em class=\"italic\">-Lowry<\/em><em class=\"italic\"> Theory what makes a compound an element or a base is whether or not it donates or accepts protons. If the H<sub>2<\/sub>O was in a different problem and was instead donating an H rather than accepting an H it would be an acid!<\/em><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"mt-section\">\n<div class=\"mt-section\">\n<h1 id=\"title\">Overview of acid-base reactions<\/h1>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignright size-thumbnail wp-image-4632\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01150925\/static_qr_code_without_logo1-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Organic acid-base mechanisms | Resonance and acid-base chemistry | Organic chemistry | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/jIL333CKE9A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"mt-section\">\n<h1 class=\"editable\">An acid-base (proton transfer) reaction<\/h1>\n<p>For our first example of chemical reactivity, let\u2019s look at a very simple reaction that occurs between hydroxide ion and hydrochloric acid:<\/p>\n<p style=\"text-align: center\">$$HCl+OH^- \\Rightarrow H_20+Cl^-$$<\/p>\n<p>This is an acid-base reaction: a proton is transferred from HCl, the acid, to hydroxide, the base. The product is water (the conjugate acid of hydroxide)\u00a0 and chloride ion (the conjugate base of HCl). You have undoubtedly seen this reaction before in general chemistry.\u00a0 Despite its simplicity (and despite the fact that the reactants and products are\u00a0<em>inorganic<\/em>\u00a0rather than organic), this reaction allows us to consider for the first time many of the fundamental ideas of organic chemistry that we will be exploring in various contexts throughout this text.<\/p>\n<p>Key to understanding just about any reaction mechanism is the concept of\u00a0<strong>electron density<\/strong>, and how it is connected to the\u00a0<strong>electron movement<\/strong>\u00a0(bond-breaking and bond-forming) that occurs in a reaction. The hydroxide ion \u2013 specifically, the electronegative oxygen atom in the hydroxide ion \u2013 has high electron density due to negative charge and the polarity of the hydrogen-oxygen bond.\u00a0\u00a0 The hydroxide oxygen is\u00a0<strong>electron-rich<\/strong>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5293\/image001.png?revision=1\" alt=\"image002.png\" width=\"187\" height=\"74\" \/><\/p>\n<p>The hydrogen atom in HCl, on the other hand, has low electron density: it is\u00a0<strong>electron-poor<\/strong>.\u00a0 As you might expect, something that is electron-rich is attracted to something that is electron-poor.\u00a0 As hydroxide and HCl move closer to each other, a lone pair of electrons on the electron-rich hydroxide oxygen is attracted by the electron-poor proton of HCl, and electron movement occurs towards the proton.\u00a0 The two electrons in the hydrogen-chlorine sigma bond are repelled by this approaching hydroxide electron density, and therefore move even farther away from the proton and towards the chlorine nucleus.\u00a0 The consequence of all of this electron movement is that the hydrogen-chlorine bond is broken, as the two electrons from that bond completely break free from the 1s orbital of the hydrogen and become a lone pair in the 3p orbital of a chloride anion.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5295\/image003.png?revision=1\" alt=\"image004.png\" width=\"369\" height=\"50\" \/><\/p>\n<p>At the same time that the hydrogen-chlorine bond is breaking, a new sigma bond forms between hydrogen and oxygen, containing the two electrons that previously were a lone pair on hydroxide.\u00a0 The result of this bond formation is, of course, a water molecule.<\/p>\n<p>Previously (<a class=\"mt-disabled\" title=\"section 2.2\" href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-1-electron-flow\/\" rel=\"broken\">section 6.1.<\/a>), we saw how curved arrows were used to depict electron movement that occurs in chemical reactions, where bonds are broken and new bonds are formed.\u00a0 The HCl + OH<sup>&#8211;<\/sup>\u00a0reaction, for example, is depicted by drawing two curved arrows.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5297\/image005.png?revision=1\" alt=\"image006.png\" width=\"321\" height=\"94\" \/><\/p>\n<p>The first arrow originates at one of the lone pairs on the hydroxide oxygen and points to the \u2018H\u2019 symbol in the hydrogen bromide molecule, illustrating the \u2018attack\u2019 of the oxygen lone pair and subsequent formation of the new hydrogen-oxygen bond.\u00a0 The second curved arrow originates at the hydrogen-bromine bond and points to the \u2018Br\u2019 symbol, indicating that this bond is breaking &#8211; the two electrons are \u2018leaving\u2019 and becoming a lone pair on bromide ion.<\/p>\n<p>It is very important to emphasize at this point that these curved,\u00a0<em>two-barbed <\/em>arrows always represent the movement of\u00a0<em>two<\/em>\u00a0electrons.\u00a0 Most of this book will be devoted to the description of reaction mechanisms involving two-electron movement, so these full-headed arrows will become very familiar.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/5299\/image007.png?revision=1\" alt=\"image008.png\" width=\"515\" height=\"95\" \/><\/p>\n<p>In the second semester, however, we will look at radical reaction mechanisms, where single-electron movement occurs.\u00a0 For these processes, a curved, single-barbed (&#8216;<strong>fish-hook<\/strong>&#8216;) arrow will be used.<\/p>\n<div>\n<div id=\"exercise\">\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<p>Draw electron movement arrows to illustrate the acid-base reaction between acetic acid, CH<sub>3<\/sub>COOH, and ammonia, NH<sub>3<\/sub>.\u00a0 Draw out the full Lewis structures of reactants and products.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q893749\">Show Solution<\/span><\/p>\n<div id=\"q893749\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/6520\/image313.png?revision=1\" alt=\"image312.png\" width=\"490\" height=\"93\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2536\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>1.8.2. Bronsted-Lowry Acids and Bases. <strong>Authored by<\/strong>: Sarah Rundle (UCD), Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/LibreTexts\/Purdue\/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)\/Chapter_01%3A_Introduction_and_Review\/1.8_Acids_and_Bases\/1.8.2._Bronsted-Lowry_Acids_and_Bases\">https:\/\/chem.libretexts.org\/LibreTexts\/Purdue\/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)\/Chapter_01%3A_Introduction_and_Review\/1.8_Acids_and_Bases\/1.8.2._Bronsted-Lowry_Acids_and_Bases<\/a>. <strong>Project<\/strong>: Chemistry LibreTexts. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><li>7.1: Overview of acid-base reactions. <strong>Authored by<\/strong>: u00a0Tim Soderbergu00a0(University of Minnesota, Morris). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_07%3A_Organic_compounds_as_acids_and_bases\/7.1%3A_Overview_of_acid-base_reactions\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_07%3A_Organic_compounds_as_acids_and_bases\/7.1%3A_Overview_of_acid-base_reactions<\/a>. <strong>Project<\/strong>: Chemistry LibreTexts. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><li>Organic Chemistry With a Biological Emphasis. <strong>Authored by<\/strong>: Tim Soderberg (University of Minnesota, Morris). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_06%3A_Introduction_to_organic_reactivity_and_catalysis\/6.1%3A_A_first_look_at_reaction_mechanisms#6.1A:_An_acid-base_(proton_transfer)_reaction\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_06%3A_Introduction_to_organic_reactivity_and_catalysis\/6.1%3A_A_first_look_at_reaction_mechanisms#6.1A:_An_acid-base_(proton_transfer)_reaction<\/a>. <strong>Project<\/strong>: Chemistry LibreTexts. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"1.8.2. 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