{"id":7629,"date":"2021-06-08T21:57:05","date_gmt":"2021-06-08T21:57:05","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/bronsted-lowry-acids-and-bases\/"},"modified":"2021-10-06T23:03:12","modified_gmt":"2021-10-06T23:03:12","slug":"bronsted-lowry-acids-and-bases","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/bronsted-lowry-acids-and-bases\/","title":{"raw":"Br\u00f8nsted-Lowry Acids and Bases","rendered":"Br\u00f8nsted-Lowry Acids and Bases"},"content":{"raw":"[latexpage]\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Identify a Br\u00f8nsted-Lowry acid and a Br\u00f8nsted-Lowry base.<\/li>\r\n \t<li>Identify conjugate acid-base pairs in an acid-base reaction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nThe Arrhenius definition of acid and base is limited to aqueous (that is, water) solutions. Although this is useful because water is a common solvent, it is limited to the relationship between the H<sup>+<\/sup> ion and the OH<sup>\u2212<\/sup> ion. What would be useful is a more general definition that would be more applicable to other chemical reactions and, importantly, independent of H<sub>2<\/sub>O.\r\n\r\nIn 1923, Danish chemist Johannes Br\u00f8nsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A <strong>Br\u00f8nsted-Lowry acid<\/strong>\u00a0is any species that can donate a proton (H<sup>+<\/sup>) to another molecule. A <strong>Br\u00f8nsted-Lowry base<\/strong>\u00a0is any species that can accept a proton from another molecule. In short, a Br\u00f8nsted-Lowry acid is a proton donor (PD), while a Br\u00f8nsted-Lowry base is a proton acceptor (PA).\r\n\r\nIt is easy to see that the Br\u00f8nsted-Lowry definition covers the Arrhenius definition of acids and bases. Consider the prototypical Arrhenius acid-base reaction:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\ce{H+(aq)}&amp;+&amp;\\ce{OH-(aq)}&amp;\\rightarrow&amp;\\ce{H2O(\\ell)} \\\\ \\\\\r\n\\text{(acid)}&amp;&amp;\\text{(base)}&amp;&amp;\r\n\\end{array}[\/latex]<\/p>\r\nThe acid species and base species are marked. The proton, however, is (by definition) a proton donor (labelled PD), while the OH<sup>\u2212<\/sup> ion is acting as the proton acceptor (labelled PA):\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\ce{H+(aq)}&amp;+&amp;\\ce{OH-(aq)}&amp;\\rightarrow&amp;\\ce{H2O(\\ell)} \\\\ \\\\\r\n\\text{(PD)}&amp;&amp;\\text{(PA)}&amp;&amp;\r\n\\end{array}[\/latex]<\/p>\r\nThe proton donor is a Br\u00f8nsted-Lowry acid, and the proton acceptor is the Br\u00f8nsted-Lowry base:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\ce{H+(aq)}&amp;+&amp;\\ce{OH-(aq)}&amp;\\rightarrow&amp;\\ce{H2O(\\ell)} \\\\ \\\\\r\n\\text{(BL acid)}&amp;&amp;\\text{(BL base)}&amp;&amp;\r\n\\end{array}[\/latex]<\/p>\r\nThus H<sup>+<\/sup> is an acid by both definitions, and OH<sup>\u2212<\/sup> is a base by both definitions.\r\n\r\nAmmonia (NH<sub>3<\/sub>) is a base even though it does not contain OH<sup>\u2212<\/sup> ions in its formula. Instead, it generates OH<sup>\u2212<\/sup> ions as the product of a proton-transfer reaction with H<sub>2<\/sub>O molecules; NH<sub>3<\/sub> acts like a Br\u00f8nsted-Lowry base, and H<sub>2<\/sub>O acts like a Br\u00f8nsted-Lowry acid:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\chemfig{H-\\Lewis{6:,N}(-[:90]H)-H}&amp;+&amp;\\chemfig{H-[:30]O-[:-30]H}&amp;\\rightarrow&amp;\\chemfig{H-N^+(-[:90]H)(-[:-90]H)-H}+\\chemfig{^{-}O-H} \\\\\r\n\\text{proton acceptor}&amp;&amp;\\text{proton donor}&amp;&amp;\r\n\\end{array}[\/latex]<\/p>\r\nA reaction with water is called <strong>hydrolysis<\/strong>; we say that NH<sub>3<\/sub> hydrolyzes to make NH<sub>4<\/sub><sup>+<\/sup> ions and OH<sup>\u2212<\/sup> ions.\r\n\r\nEven the dissolving of an Arrhenius acid in water can be considered a Br\u00f8nsted-Lowry acid-base reaction. Consider the process of dissolving HCl(g) in water to make an aqueous solution of hydrochloric acid. The process can be written as follows:\r\n<p style=\"text-align: center;\">HCl(g) +\u00a0H<sub>2<\/sub>O(\u2113) \u2192\u00a0H<sub>3<\/sub>O<sup>+<\/sup>(aq) +\u00a0Cl<sup>\u2212<\/sup>(aq)<\/p>\r\nHCl(g) is the proton donor and therefore a Br\u00f8nsted-Lowry acid, while H<sub>2<\/sub>O is the proton acceptor and a Br\u00f8nsted-Lowry base. These two examples show that H<sub>2<\/sub>O can act as both a proton donor and a proton acceptor, depending on what other substance is in the chemical reaction. A substance that can act as a proton donor or a proton acceptor is called <strong>amphiprotic<\/strong>. Water is probably the most common amphiprotic substance we will encounter, but other substances are also amphiprotic.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIdentify the Br\u00f8nsted-Lowry acid and the Br\u00f8nsted-Lowry base in this chemical equation.\r\n<p style=\"text-align: center;\">C<sub>6<\/sub>H<sub>5<\/sub>OH + NH<sub>2<\/sub><sup>\u2212<\/sup> \u2192 C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> + NH<sub>3<\/sub><\/p>\r\n<em>Solution<\/em>\r\nThe C<sub>6<\/sub>H<sub>5<\/sub>OH molecule is losing an H<sup>+<\/sup>; it is the proton donor and the Br\u00f8nsted-Lowry acid. The NH<sub>2<\/sub><sup>\u2212<\/sup> ion (called the amide ion) is accepting the H<sup>+<\/sup> ion to become NH<sub>3<\/sub>, so it is the Br\u00f8nsted-Lowry base.\r\n\r\n<em>Test Yourself<\/em>\r\nIdentify the Br\u00f8nsted-Lowry acid and the Br\u00f8nsted-Lowry base in this chemical equation.\r\n<p style=\"text-align: center;\">Al(H<sub>2<\/sub>O)<sub>6<\/sub><sup>3+<\/sup> + H<sub>2<\/sub>O \u2192 Al(H<sub>2<\/sub>O)<sub>5<\/sub>(OH)<sup>2+<\/sup> + H<sub>3<\/sub>O<sup>+<\/sup><\/p>\r\n<em>Answer<\/em>\r\nBr\u00f8nsted-Lowry acid: Al(H<sub>2<\/sub>O)<sub>6<\/sub><sup>3+<\/sup>; Br\u00f8nsted-Lowry base: H<sub>2<\/sub>O\r\n\r\n<\/div>\r\n<\/div>\r\n<p class=\"page-break-before\">Observe the reaction between NH<sub>3<\/sub> and H<sub>2<\/sub>O:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\chemfig{H-\\Lewis{6:,N}(-[:90]H)-H}&amp;+&amp;\\chemfig{H-[:30]O-[:-30]H} &amp;\\rightarrow &amp; \\chemfig{H-N^+(-[:90]H)(-[:-90]H)-H}+\\chemfig{O-H^{+}}}\r\n\\end{array}[\/latex]<\/p>\r\nThe chemical reaction does not go to completion; rather, the reverse process occurs as well, and eventually the two processes cancel out any additional change. At this point, we say the chemical reaction is at <em>equilibrium<\/em>. Both processes still occur, but any net change by one process is countered by the same net change by the other process; it is a <em>dynamic<\/em>, rather than a <em>static<\/em>, equilibrium. Because both reactions are occurring, it makes sense to use a double arrow instead of a single arrow:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\chemfig{H-\\Lewis{6:,N}(-[:90]H)-H}&amp;+&amp;\\chemfig{H-[:30]O-[:-30]H} &amp;\\rightleftharpoons &amp; \\chemfig{H-N^+(-[:90]H)(-[:-90]H)-H}+\\chemfig{O-H^{-}}}\r\n\\end{array}[\/latex]<\/p>\r\nWhat do you notice about the reverse reaction? The NH<sub>4<\/sub><sup>+<\/sup> ion is donating a proton to the OH<sup>\u2212<\/sup> ion, which is accepting it. This means that the NH<sub>4<\/sub><sup>+<\/sup> ion is acting as the proton donor, or Br\u00f8nsted-Lowry acid, while OH<sup>\u2212<\/sup> ion, the proton acceptor, is acting as a Br\u00f8nsted-Lowry base. The reverse reaction is also a Br\u00f8nsted-Lowry acid base reaction:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccccc}\r\n\\chemfig{H-\\Lewis{6:,N}(-[:90]H)-H}&amp;+&amp;\\chemfig{H-[:30]O-[:-30]H} &amp;\\rightleftharpoons &amp; \\chemfig{H-N^+(-[:90]H)(-[:-90]H)-H}+\\chemfig{O-H^{-}}}\r\n\\end{array}[\/latex]<\/p>\r\nThis means that both reactions are acid-base reactions by the Br\u00f8nsted-Lowry definition. If you consider the species in this chemical reaction, two sets of similar species exist on both sides. Within each set, the two species differ by a proton in their formulas, and one member of the set is a Br\u00f8nsted-Lowry acid, while the other member is a Br\u00f8nsted-Lowry base. These sets are marked here:\r\n\r\n<img class=\"size-full wp-image-685 aligncenter\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/BL-Acid-Base-Reaction-2-1.png\" alt=\"BL Acid-Base Reaction 2\" width=\"600\" height=\"170\" \/>\r\n\r\nThe two sets \u2014 NH<sub>3<\/sub>\/NH<sub>4<\/sub><sup>+<\/sup> and H<sub>2<\/sub>O\/OH<sup>\u2212<\/sup> \u2014 are called <strong>conjugate acid-base pairs<\/strong>. We say that NH<sub>4<\/sub><sup>+<\/sup> is the conjugate acid of NH<sub>3<\/sub>, OH<sup>\u2212<\/sup> is the conjugate base of H<sub>2<\/sub>O, and so forth. Every Br\u00f8nsted-Lowry acid-base reaction can be labelled with two conjugate acid-base pairs.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 12.6<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIdentify the conjugate acid-base pairs in this equilibrium.\r\n<p style=\"text-align: center;\">(CH<sub>3<\/sub>)<sub>3<\/sub>N + H<sub>2<\/sub>O \u21c4 (CH<sub>3<\/sub>)<sub>3<\/sub>NH+ + OH<sup>\u2013<\/sup><\/p>\r\n<em>Solution<\/em>\r\nOne pair is H<sub>2<\/sub>O and OH<sup>\u2212<\/sup>, where H<sub>2<\/sub>O has one more H<sup>+<\/sup> and is the conjugate acid, while OH<sup>\u2212<\/sup> has one less H<sup>+<\/sup> and is the conjugate base. The other pair consists of (CH<sub>3<\/sub>)<sub>3<\/sub>N and (CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup>, where (CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup> is the conjugate acid (it has an additional proton) and (CH<sub>3<\/sub>)<sub>3<\/sub>N is the conjugate base.\r\n\r\n<em>Test Yourself<\/em>\r\nIdentify the conjugate acid-base pairs in this equilibrium.\r\n<p style=\"text-align: center;\">NH<sub>2<\/sub><sup>\u2013 <\/sup>+ H<sub>2<\/sub>O \u21c4 NH<sub>3<\/sub> + OH<sup>\u2013<\/sup><\/p>\r\n<em>Answer<\/em>\r\nH<sub>2<\/sub>O (acid) and OH<sup>\u2212<\/sup> (base); NH<sub>2<\/sub><sup>\u2212<\/sup> (base) and NH<sub>3<\/sub> (acid)\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h1>Chemistry Is Everywhere: Household Acids and Bases<\/h1>\r\nMany household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for HCl(aq). In <a class=\"internal\" href=\"\/introductorychemistry\/part\/chapter-4-chemical-reactions-and-equations\/\">Chapter 4 \"Chemical Reactions and Equations\"<\/a>, in the section <a class=\"internal\" href=\"\/introductorychemistry\/chapter\/neutralization-reactions\/\">\"Neutralization Reactions\"<\/a>, vinegar was mentioned as a dilute solution of acetic acid [HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>(aq)]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid (HC<sub>6<\/sub>H<sub>7<\/sub>O<sub>6<\/sub>).\r\n\r\nOne of the more familiar household bases is NH<sub>3<\/sub>, which is found in numerous cleaning products. NH<sub>3<\/sub> is a base because it increases the OH<sup>\u2212<\/sup> ion concentration by reacting with H<sub>2<\/sub>O:\r\n<p style=\"text-align: center;\">NH<sub>3<\/sub>(aq) +\u00a0H<sub>2<\/sub>O(\u2113) \u2192\u00a0NH<sub>4<\/sub><sup>+<\/sup>(aq) +\u00a0OH<sup>\u2212<\/sup>(aq)<\/p>\r\nMany soaps are also slightly basic because they contain compounds that act as Br\u00f8nsted-Lowry bases, accepting protons from H<sub>2<\/sub>O and forming excess OH<sup>\u2212<\/sup> ions. This is one explanation for why soap solutions are slippery.\r\n\r\nPerhaps the most dangerous household chemical is the lye-based drain cleaner. Lye is a common name for NaOH, although it is also used as a synonym for KOH. Lye is an extremely caustic chemical that can react with grease, hair, food particles, and other substances that may build up and clog a water pipe. Unfortunately, lye can also attack body tissues and other substances in our bodies. Thus when we use lye-based drain cleaners, we must be very careful not to touch any of the solid drain cleaner or spill the water it was poured into. Safer, nonlye drain cleaners (like the one in the accompanying figure) use peroxide compounds to react on the materials in the clog and clear the drain.\r\n\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li>A Br\u00f8nsted-Lowry acid is a proton donor; a Br\u00f8nsted-Lowry base is a proton acceptor.<\/li>\r\n \t<li>Acid-base reactions include two sets of conjugate acid-base pairs.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercises<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<h1>Questions<\/h1>\r\n<ol>\r\n \t<li>Define <em>Br\u00f8nsted-Lowry acid<\/em>. How does it differ from an Arrhenius acid?<\/li>\r\n \t<li>Define <em>Br\u00f8nsted-Lowry base<\/em>. How does it differ from an Arrhenius base?<\/li>\r\n \t<li>Write the dissociation of hydrogen bromide in water as a Br\u00f8nsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.<\/li>\r\n \t<li>Write the dissociation of nitric acid in water as a Br\u00f8nsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.<\/li>\r\n \t<li>Pyridine (C<sub>5<\/sub>H<sub>5<\/sub>N) acts as a Br\u00f8nsted-Lowry base in water. Write the hydrolysis reaction for pyridine and identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base.<\/li>\r\n \t<li>The methoxide ion (CH<sub>3<\/sub>O<sup>\u2212<\/sup>) acts as a Br\u00f8nsted-Lowry base in water. Write the hydrolysis reaction for the methoxide ion and identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base.<\/li>\r\n \t<li>Identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base in this chemical equation.\r\n<p style=\"text-align: center;\">H<sub>3<\/sub>PO<sub>4<\/sub> +\u00a0OH<sup>\u2212<\/sup> \u2192\u00a0H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup> +\u00a0H<sub>2<\/sub>O<\/p>\r\n<\/li>\r\n \t<li>Identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base in this chemical equation.\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> +\u00a02F<sup>\u2212<\/sup> \u2192\u00a02HF +\u00a0C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup><\/p>\r\n<\/li>\r\n \t<li>Predict the products of this reaction, assuming it undergoes a Br\u00f8nsted-Lowry acid-base reaction.\r\n<p style=\"text-align: center;\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub> +\u00a0C<sub>5<\/sub>H<sub>5<\/sub>N \u2192\u00a0?<\/p>\r\n<\/li>\r\n \t<li>Predict the products of this reaction, assuming it undergoes a Br\u00f8nsted-Lowry acid-base reaction.\r\n<p style=\"text-align: center;\">(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>N +\u00a0H<sub>2<\/sub>O \u2192\u00a0?<\/p>\r\n<\/li>\r\n \t<li>What is the conjugate acid of H<sub>2<\/sub>O? NH<sub>3<\/sub>?<\/li>\r\n \t<li>What is the conjugate acid of H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>? NO<sub>3<\/sub><sup>\u2212<\/sup>?<\/li>\r\n \t<li>What is the conjugate base of HSO<sub>4<\/sub><sup>\u2212<\/sup>? H<sub>2<\/sub>O?<\/li>\r\n \t<li>What is the conjugate base of H<sub>3<\/sub>O<sup>+<\/sup>? H<sub>2<\/sub>SO<sub>4<\/sub>?<\/li>\r\n \t<li>Identify the conjugate acid-base pairs in this reaction.\r\n<p style=\"text-align: center;\">HSO<sub>4<\/sub><sup>\u2212<\/sup> +\u00a0PO<sub>4<\/sub><sup>3\u2212<\/sup> \u2192\u00a0SO<sub>4<\/sub><sup>2\u2212<\/sup> +\u00a0HPO<sub>4<\/sub><sup>2\u2212<\/sup><\/p>\r\n<\/li>\r\n \t<li>Identify the conjugate acid-base pairs in this reaction.\r\n<p style=\"text-align: center;\">HClO<sub>3<\/sub> +\u00a0(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>N \u2192\u00a0ClO<sub>3<\/sub><sup>\u2212<\/sup> +\u00a0(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup><\/p>\r\n<\/li>\r\n \t<li>Identify the conjugate acid-base pairs in this reaction.\r\n<p style=\"text-align: center;\">NH<sub>3<\/sub> +\u00a0C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> \u2192\u00a0C<sub>6<\/sub>H<sub>5<\/sub>OH +\u00a0NH<sub>2<\/sub><sup>\u2212<\/sup><\/p>\r\n<\/li>\r\n \t<li>Identify the conjugate acid-base pairs in this reaction.\r\n<p style=\"text-align: center;\">C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup> +\u00a0C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup> \u2192\u00a0C<sub>5<\/sub>H<sub>5<\/sub>N +\u00a0HC<sub>2<\/sub>O<sub>4<\/sub><sup>\u2212<\/sup><\/p>\r\n<\/li>\r\n<\/ol>\r\n<h1>Answers<\/h1>\r\n<ol>\r\n \t<li>A Br\u00f8nsted-Lowry acid is a proton donor. It does not necessarily increase the H<sup>+<\/sup> concentration in water.<\/li>\r\n<\/ol>\r\n<ol start=\"3\">\r\n \t<li>HBr +\u00a0H<sub>2<\/sub>O \u2192\u00a0H<sub>3<\/sub>O<sup>+<\/sup> +\u00a0Br<sup>\u2212<\/sup>; PD: HBr; PA: H<sub>2<\/sub>O<\/li>\r\n<\/ol>\r\n<ol start=\"5\">\r\n \t<li>C<sub>5<\/sub>H<sub>5<\/sub>N +\u00a0H<sub>2<\/sub>O \u2192\u00a0C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup> +\u00a0OH<sup>\u2212<\/sup>; PD: H<sub>2<\/sub>O; PA: C<sub>5<\/sub>H<sub>5<\/sub>N<\/li>\r\n<\/ol>\r\n<ol start=\"7\">\r\n \t<li>BL acid: H<sub>3<\/sub>PO<sub>4<\/sub>; BL base: OH<sup>\u2212<\/sup><\/li>\r\n<\/ol>\r\n<ol start=\"9\">\r\n \t<li>C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup> and C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup><\/li>\r\n<\/ol>\r\n<ol start=\"11\">\r\n \t<li>H<sub>3<\/sub>O<sup>+<\/sup>; NH<sub>4<\/sub><sup>+<\/sup><\/li>\r\n<\/ol>\r\n<ol start=\"13\">\r\n \t<li>SO<sub>4<\/sub><sup>2\u2212<\/sup>; OH<sup>\u2212<\/sup><\/li>\r\n<\/ol>\r\n<ol start=\"15\">\r\n \t<li>HSO<sub>4<\/sub><sup>\u2212<\/sup> and SO<sub>4<\/sub><sup>2\u2212<\/sup>; PO<sub>4<\/sub><sup>3\u2212<\/sup> and HPO<sub>4<\/sub><sup>2\u2212<\/sup><\/li>\r\n<\/ol>\r\n<ol start=\"17\">\r\n \t<li>NH<sub>3<\/sub> and NH<sub>2<\/sub><sup>\u2212<\/sup>; C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> and C<sub>6<\/sub>H<sub>5<\/sub>OH<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Identify a Br\u00f8nsted-Lowry acid and a Br\u00f8nsted-Lowry base.<\/li>\n<li>Identify conjugate acid-base pairs in an acid-base reaction.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>The Arrhenius definition of acid and base is limited to aqueous (that is, water) solutions. Although this is useful because water is a common solvent, it is limited to the relationship between the H<sup>+<\/sup> ion and the OH<sup>\u2212<\/sup> ion. What would be useful is a more general definition that would be more applicable to other chemical reactions and, importantly, independent of H<sub>2<\/sub>O.<\/p>\n<p>In 1923, Danish chemist Johannes Br\u00f8nsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A <strong>Br\u00f8nsted-Lowry acid<\/strong>\u00a0is any species that can donate a proton (H<sup>+<\/sup>) to another molecule. A <strong>Br\u00f8nsted-Lowry base<\/strong>\u00a0is any species that can accept a proton from another molecule. In short, a Br\u00f8nsted-Lowry acid is a proton donor (PD), while a Br\u00f8nsted-Lowry base is a proton acceptor (PA).<\/p>\n<p>It is easy to see that the Br\u00f8nsted-Lowry definition covers the Arrhenius definition of acids and bases. Consider the prototypical Arrhenius acid-base reaction:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-c282728d76e3798f016500f912df5d03_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#101;&#123;&#72;&#43;&#40;&#97;&#113;&#41;&#125;&#38;&#43;&#38;&#92;&#99;&#101;&#123;&#79;&#72;&#45;&#40;&#97;&#113;&#41;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#99;&#101;&#123;&#72;&#50;&#79;&#40;&#92;&#101;&#108;&#108;&#41;&#125;&#32;&#92;&#92;&#32;&#92;&#92; &#92;&#116;&#101;&#120;&#116;&#123;&#40;&#97;&#99;&#105;&#100;&#41;&#125;&#38;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#40;&#98;&#97;&#115;&#101;&#41;&#125;&#38;&#38; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"64\" width=\"279\" style=\"vertical-align: -27px;\" \/><\/p>\n<p>The acid species and base species are marked. The proton, however, is (by definition) a proton donor (labelled PD), while the OH<sup>\u2212<\/sup> ion is acting as the proton acceptor (labelled PA):<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-e829399f71bb69d2bf6e7ad95d434d54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#101;&#123;&#72;&#43;&#40;&#97;&#113;&#41;&#125;&#38;&#43;&#38;&#92;&#99;&#101;&#123;&#79;&#72;&#45;&#40;&#97;&#113;&#41;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#99;&#101;&#123;&#72;&#50;&#79;&#40;&#92;&#101;&#108;&#108;&#41;&#125;&#32;&#92;&#92;&#32;&#92;&#92; &#92;&#116;&#101;&#120;&#116;&#123;&#40;&#80;&#68;&#41;&#125;&#38;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#40;&#80;&#65;&#41;&#125;&#38;&#38; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"64\" width=\"279\" style=\"vertical-align: -27px;\" \/><\/p>\n<p>The proton donor is a Br\u00f8nsted-Lowry acid, and the proton acceptor is the Br\u00f8nsted-Lowry base:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-b7c8fa535d6d8148a1c6e2b428a37c6c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#101;&#123;&#72;&#43;&#40;&#97;&#113;&#41;&#125;&#38;&#43;&#38;&#92;&#99;&#101;&#123;&#79;&#72;&#45;&#40;&#97;&#113;&#41;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#99;&#101;&#123;&#72;&#50;&#79;&#40;&#92;&#101;&#108;&#108;&#41;&#125;&#32;&#92;&#92;&#32;&#92;&#92; &#92;&#116;&#101;&#120;&#116;&#123;&#40;&#66;&#76;&#32;&#97;&#99;&#105;&#100;&#41;&#125;&#38;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#40;&#66;&#76;&#32;&#98;&#97;&#115;&#101;&#41;&#125;&#38;&#38; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"64\" width=\"302\" style=\"vertical-align: -27px;\" \/><\/p>\n<p>Thus H<sup>+<\/sup> is an acid by both definitions, and OH<sup>\u2212<\/sup> is a base by both definitions.<\/p>\n<p>Ammonia (NH<sub>3<\/sub>) is a base even though it does not contain OH<sup>\u2212<\/sup> ions in its formula. Instead, it generates OH<sup>\u2212<\/sup> ions as the product of a proton-transfer reaction with H<sub>2<\/sub>O molecules; NH<sub>3<\/sub> acts like a Br\u00f8nsted-Lowry base, and H<sub>2<\/sub>O acts like a Br\u00f8nsted-Lowry acid:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-1c9f61fcf46dc1879138361cc4a245c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#92;&#76;&#101;&#119;&#105;&#115;&#123;&#54;&#58;&#44;&#78;&#125;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#38;&#43;&#38;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#91;&#58;&#51;&#48;&#93;&#79;&#45;&#91;&#58;&#45;&#51;&#48;&#93;&#72;&#125;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#38;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#78;&#94;&#43;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#40;&#45;&#91;&#58;&#45;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#43;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#94;&#123;&#45;&#125;&#79;&#45;&#72;&#125;&#32;&#92;&#92; &#92;&#116;&#101;&#120;&#116;&#123;&#112;&#114;&#111;&#116;&#111;&#110;&#32;&#97;&#99;&#99;&#101;&#112;&#116;&#111;&#114;&#125;&#38;&#38;&#92;&#116;&#101;&#120;&#116;&#123;&#112;&#114;&#111;&#116;&#111;&#110;&#32;&#100;&#111;&#110;&#111;&#114;&#125;&#38;&#38; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"141\" width=\"543\" style=\"vertical-align: 0px;\" \/><\/p>\n<p>A reaction with water is called <strong>hydrolysis<\/strong>; we say that NH<sub>3<\/sub> hydrolyzes to make NH<sub>4<\/sub><sup>+<\/sup> ions and OH<sup>\u2212<\/sup> ions.<\/p>\n<p>Even the dissolving of an Arrhenius acid in water can be considered a Br\u00f8nsted-Lowry acid-base reaction. Consider the process of dissolving HCl(g) in water to make an aqueous solution of hydrochloric acid. The process can be written as follows:<\/p>\n<p style=\"text-align: center;\">HCl(g) +\u00a0H<sub>2<\/sub>O(\u2113) \u2192\u00a0H<sub>3<\/sub>O<sup>+<\/sup>(aq) +\u00a0Cl<sup>\u2212<\/sup>(aq)<\/p>\n<p>HCl(g) is the proton donor and therefore a Br\u00f8nsted-Lowry acid, while H<sub>2<\/sub>O is the proton acceptor and a Br\u00f8nsted-Lowry base. These two examples show that H<sub>2<\/sub>O can act as both a proton donor and a proton acceptor, depending on what other substance is in the chemical reaction. A substance that can act as a proton donor or a proton acceptor is called <strong>amphiprotic<\/strong>. Water is probably the most common amphiprotic substance we will encounter, but other substances are also amphiprotic.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Identify the Br\u00f8nsted-Lowry acid and the Br\u00f8nsted-Lowry base in this chemical equation.<\/p>\n<p style=\"text-align: center;\">C<sub>6<\/sub>H<sub>5<\/sub>OH + NH<sub>2<\/sub><sup>\u2212<\/sup> \u2192 C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> + NH<sub>3<\/sub><\/p>\n<p><em>Solution<\/em><br \/>\nThe C<sub>6<\/sub>H<sub>5<\/sub>OH molecule is losing an H<sup>+<\/sup>; it is the proton donor and the Br\u00f8nsted-Lowry acid. The NH<sub>2<\/sub><sup>\u2212<\/sup> ion (called the amide ion) is accepting the H<sup>+<\/sup> ion to become NH<sub>3<\/sub>, so it is the Br\u00f8nsted-Lowry base.<\/p>\n<p><em>Test Yourself<\/em><br \/>\nIdentify the Br\u00f8nsted-Lowry acid and the Br\u00f8nsted-Lowry base in this chemical equation.<\/p>\n<p style=\"text-align: center;\">Al(H<sub>2<\/sub>O)<sub>6<\/sub><sup>3+<\/sup> + H<sub>2<\/sub>O \u2192 Al(H<sub>2<\/sub>O)<sub>5<\/sub>(OH)<sup>2+<\/sup> + H<sub>3<\/sub>O<sup>+<\/sup><\/p>\n<p><em>Answer<\/em><br \/>\nBr\u00f8nsted-Lowry acid: Al(H<sub>2<\/sub>O)<sub>6<\/sub><sup>3+<\/sup>; Br\u00f8nsted-Lowry base: H<sub>2<\/sub>O<\/p>\n<\/div>\n<\/div>\n<p class=\"page-break-before\">Observe the reaction between NH<sub>3<\/sub> and H<sub>2<\/sub>O:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-642d27998cd84cd45168ed40d6abdc6f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#92;&#76;&#101;&#119;&#105;&#115;&#123;&#54;&#58;&#44;&#78;&#125;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#38;&#43;&#38;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#91;&#58;&#51;&#48;&#93;&#79;&#45;&#91;&#58;&#45;&#51;&#48;&#93;&#72;&#125;&#32;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#97;&#114;&#114;&#111;&#119;&#32;&#38;&#32;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#78;&#94;&#43;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#40;&#45;&#91;&#58;&#45;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#43;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#79;&#45;&#72;&#94;&#123;&#43;&#125;&#125;&#125; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"122\" width=\"533\" style=\"vertical-align: 0px;\" \/><\/p>\n<p>The chemical reaction does not go to completion; rather, the reverse process occurs as well, and eventually the two processes cancel out any additional change. At this point, we say the chemical reaction is at <em>equilibrium<\/em>. Both processes still occur, but any net change by one process is countered by the same net change by the other process; it is a <em>dynamic<\/em>, rather than a <em>static<\/em>, equilibrium. Because both reactions are occurring, it makes sense to use a double arrow instead of a single arrow:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-12aa9f53144534395d55fdf4058723fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#92;&#76;&#101;&#119;&#105;&#115;&#123;&#54;&#58;&#44;&#78;&#125;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#38;&#43;&#38;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#91;&#58;&#51;&#48;&#93;&#79;&#45;&#91;&#58;&#45;&#51;&#48;&#93;&#72;&#125;&#32;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#108;&#101;&#102;&#116;&#104;&#97;&#114;&#112;&#111;&#111;&#110;&#115;&#32;&#38;&#32;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#78;&#94;&#43;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#40;&#45;&#91;&#58;&#45;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#43;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#79;&#45;&#72;&#94;&#123;&#45;&#125;&#125;&#125; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"122\" width=\"533\" style=\"vertical-align: 0px;\" \/><\/p>\n<p>What do you notice about the reverse reaction? The NH<sub>4<\/sub><sup>+<\/sup> ion is donating a proton to the OH<sup>\u2212<\/sup> ion, which is accepting it. This means that the NH<sub>4<\/sub><sup>+<\/sup> ion is acting as the proton donor, or Br\u00f8nsted-Lowry acid, while OH<sup>\u2212<\/sup> ion, the proton acceptor, is acting as a Br\u00f8nsted-Lowry base. The reverse reaction is also a Br\u00f8nsted-Lowry acid base reaction:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/ql-cache\/quicklatex.com-12aa9f53144534395d55fdf4058723fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#125; &#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#92;&#76;&#101;&#119;&#105;&#115;&#123;&#54;&#58;&#44;&#78;&#125;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#38;&#43;&#38;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#91;&#58;&#51;&#48;&#93;&#79;&#45;&#91;&#58;&#45;&#51;&#48;&#93;&#72;&#125;&#32;&#38;&#92;&#114;&#105;&#103;&#104;&#116;&#108;&#101;&#102;&#116;&#104;&#97;&#114;&#112;&#111;&#111;&#110;&#115;&#32;&#38;&#32;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#72;&#45;&#78;&#94;&#43;&#40;&#45;&#91;&#58;&#57;&#48;&#93;&#72;&#41;&#40;&#45;&#91;&#58;&#45;&#57;&#48;&#93;&#72;&#41;&#45;&#72;&#125;&#43;&#92;&#99;&#104;&#101;&#109;&#102;&#105;&#103;&#123;&#79;&#45;&#72;&#94;&#123;&#45;&#125;&#125;&#125; &#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"122\" width=\"533\" style=\"vertical-align: 0px;\" \/><\/p>\n<p>This means that both reactions are acid-base reactions by the Br\u00f8nsted-Lowry definition. If you consider the species in this chemical reaction, two sets of similar species exist on both sides. Within each set, the two species differ by a proton in their formulas, and one member of the set is a Br\u00f8nsted-Lowry acid, while the other member is a Br\u00f8nsted-Lowry base. These sets are marked here:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-685 aligncenter\" src=\"https:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2021\/06\/BL-Acid-Base-Reaction-2-1.png\" alt=\"BL Acid-Base Reaction 2\" width=\"600\" height=\"170\" \/><\/p>\n<p>The two sets \u2014 NH<sub>3<\/sub>\/NH<sub>4<\/sub><sup>+<\/sup> and H<sub>2<\/sub>O\/OH<sup>\u2212<\/sup> \u2014 are called <strong>conjugate acid-base pairs<\/strong>. We say that NH<sub>4<\/sub><sup>+<\/sup> is the conjugate acid of NH<sub>3<\/sub>, OH<sup>\u2212<\/sup> is the conjugate base of H<sub>2<\/sub>O, and so forth. Every Br\u00f8nsted-Lowry acid-base reaction can be labelled with two conjugate acid-base pairs.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 12.6<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Identify the conjugate acid-base pairs in this equilibrium.<\/p>\n<p style=\"text-align: center;\">(CH<sub>3<\/sub>)<sub>3<\/sub>N + H<sub>2<\/sub>O \u21c4 (CH<sub>3<\/sub>)<sub>3<\/sub>NH+ + OH<sup>\u2013<\/sup><\/p>\n<p><em>Solution<\/em><br \/>\nOne pair is H<sub>2<\/sub>O and OH<sup>\u2212<\/sup>, where H<sub>2<\/sub>O has one more H<sup>+<\/sup> and is the conjugate acid, while OH<sup>\u2212<\/sup> has one less H<sup>+<\/sup> and is the conjugate base. The other pair consists of (CH<sub>3<\/sub>)<sub>3<\/sub>N and (CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup>, where (CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup> is the conjugate acid (it has an additional proton) and (CH<sub>3<\/sub>)<sub>3<\/sub>N is the conjugate base.<\/p>\n<p><em>Test Yourself<\/em><br \/>\nIdentify the conjugate acid-base pairs in this equilibrium.<\/p>\n<p style=\"text-align: center;\">NH<sub>2<\/sub><sup>\u2013 <\/sup>+ H<sub>2<\/sub>O \u21c4 NH<sub>3<\/sub> + OH<sup>\u2013<\/sup><\/p>\n<p><em>Answer<\/em><br \/>\nH<sub>2<\/sub>O (acid) and OH<sup>\u2212<\/sup> (base); NH<sub>2<\/sub><sup>\u2212<\/sup> (base) and NH<sub>3<\/sub> (acid)<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h1>Chemistry Is Everywhere: Household Acids and Bases<\/h1>\n<p>Many household products are acids or bases. For example, the owner of a swimming pool may use muriatic acid to clean the pool. Muriatic acid is another name for HCl(aq). In <a class=\"internal\" href=\"\/introductorychemistry\/part\/chapter-4-chemical-reactions-and-equations\/\">Chapter 4 &#8220;Chemical Reactions and Equations&#8221;<\/a>, in the section <a class=\"internal\" href=\"\/introductorychemistry\/chapter\/neutralization-reactions\/\">&#8220;Neutralization Reactions&#8221;<\/a>, vinegar was mentioned as a dilute solution of acetic acid [HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>(aq)]. In a medicine chest, one may find a bottle of vitamin C tablets; the chemical name of vitamin C is ascorbic acid (HC<sub>6<\/sub>H<sub>7<\/sub>O<sub>6<\/sub>).<\/p>\n<p>One of the more familiar household bases is NH<sub>3<\/sub>, which is found in numerous cleaning products. NH<sub>3<\/sub> is a base because it increases the OH<sup>\u2212<\/sup> ion concentration by reacting with H<sub>2<\/sub>O:<\/p>\n<p style=\"text-align: center;\">NH<sub>3<\/sub>(aq) +\u00a0H<sub>2<\/sub>O(\u2113) \u2192\u00a0NH<sub>4<\/sub><sup>+<\/sup>(aq) +\u00a0OH<sup>\u2212<\/sup>(aq)<\/p>\n<p>Many soaps are also slightly basic because they contain compounds that act as Br\u00f8nsted-Lowry bases, accepting protons from H<sub>2<\/sub>O and forming excess OH<sup>\u2212<\/sup> ions. This is one explanation for why soap solutions are slippery.<\/p>\n<p>Perhaps the most dangerous household chemical is the lye-based drain cleaner. Lye is a common name for NaOH, although it is also used as a synonym for KOH. Lye is an extremely caustic chemical that can react with grease, hair, food particles, and other substances that may build up and clog a water pipe. Unfortunately, lye can also attack body tissues and other substances in our bodies. Thus when we use lye-based drain cleaners, we must be very careful not to touch any of the solid drain cleaner or spill the water it was poured into. Safer, nonlye drain cleaners (like the one in the accompanying figure) use peroxide compounds to react on the materials in the clog and clear the drain.<\/p>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li>A Br\u00f8nsted-Lowry acid is a proton donor; a Br\u00f8nsted-Lowry base is a proton acceptor.<\/li>\n<li>Acid-base reactions include two sets of conjugate acid-base pairs.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercises<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<h1>Questions<\/h1>\n<ol>\n<li>Define <em>Br\u00f8nsted-Lowry acid<\/em>. How does it differ from an Arrhenius acid?<\/li>\n<li>Define <em>Br\u00f8nsted-Lowry base<\/em>. How does it differ from an Arrhenius base?<\/li>\n<li>Write the dissociation of hydrogen bromide in water as a Br\u00f8nsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.<\/li>\n<li>Write the dissociation of nitric acid in water as a Br\u00f8nsted-Lowry acid-base reaction and identify the proton donor and proton acceptor.<\/li>\n<li>Pyridine (C<sub>5<\/sub>H<sub>5<\/sub>N) acts as a Br\u00f8nsted-Lowry base in water. Write the hydrolysis reaction for pyridine and identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base.<\/li>\n<li>The methoxide ion (CH<sub>3<\/sub>O<sup>\u2212<\/sup>) acts as a Br\u00f8nsted-Lowry base in water. Write the hydrolysis reaction for the methoxide ion and identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base.<\/li>\n<li>Identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base in this chemical equation.\n<p style=\"text-align: center;\">H<sub>3<\/sub>PO<sub>4<\/sub> +\u00a0OH<sup>\u2212<\/sup> \u2192\u00a0H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup> +\u00a0H<sub>2<\/sub>O<\/p>\n<\/li>\n<li>Identify the Br\u00f8nsted-Lowry acid and Br\u00f8nsted-Lowry base in this chemical equation.\n<p style=\"text-align: center;\">H<sub>2<\/sub>C<sub>2<\/sub>O<sub>4<\/sub> +\u00a02F<sup>\u2212<\/sup> \u2192\u00a02HF +\u00a0C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup><\/p>\n<\/li>\n<li>Predict the products of this reaction, assuming it undergoes a Br\u00f8nsted-Lowry acid-base reaction.\n<p style=\"text-align: center;\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub> +\u00a0C<sub>5<\/sub>H<sub>5<\/sub>N \u2192\u00a0?<\/p>\n<\/li>\n<li>Predict the products of this reaction, assuming it undergoes a Br\u00f8nsted-Lowry acid-base reaction.\n<p style=\"text-align: center;\">(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>N +\u00a0H<sub>2<\/sub>O \u2192\u00a0?<\/p>\n<\/li>\n<li>What is the conjugate acid of H<sub>2<\/sub>O? NH<sub>3<\/sub>?<\/li>\n<li>What is the conjugate acid of H<sub>2<\/sub>PO<sub>4<\/sub><sup>\u2212<\/sup>? NO<sub>3<\/sub><sup>\u2212<\/sup>?<\/li>\n<li>What is the conjugate base of HSO<sub>4<\/sub><sup>\u2212<\/sup>? H<sub>2<\/sub>O?<\/li>\n<li>What is the conjugate base of H<sub>3<\/sub>O<sup>+<\/sup>? H<sub>2<\/sub>SO<sub>4<\/sub>?<\/li>\n<li>Identify the conjugate acid-base pairs in this reaction.\n<p style=\"text-align: center;\">HSO<sub>4<\/sub><sup>\u2212<\/sup> +\u00a0PO<sub>4<\/sub><sup>3\u2212<\/sup> \u2192\u00a0SO<sub>4<\/sub><sup>2\u2212<\/sup> +\u00a0HPO<sub>4<\/sub><sup>2\u2212<\/sup><\/p>\n<\/li>\n<li>Identify the conjugate acid-base pairs in this reaction.\n<p style=\"text-align: center;\">HClO<sub>3<\/sub> +\u00a0(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>N \u2192\u00a0ClO<sub>3<\/sub><sup>\u2212<\/sup> +\u00a0(C<sub>2<\/sub>H<sub>5<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup><\/p>\n<\/li>\n<li>Identify the conjugate acid-base pairs in this reaction.\n<p style=\"text-align: center;\">NH<sub>3<\/sub> +\u00a0C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> \u2192\u00a0C<sub>6<\/sub>H<sub>5<\/sub>OH +\u00a0NH<sub>2<\/sub><sup>\u2212<\/sup><\/p>\n<\/li>\n<li>Identify the conjugate acid-base pairs in this reaction.\n<p style=\"text-align: center;\">C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup> +\u00a0C<sub>2<\/sub>O<sub>4<\/sub><sup>2\u2212<\/sup> \u2192\u00a0C<sub>5<\/sub>H<sub>5<\/sub>N +\u00a0HC<sub>2<\/sub>O<sub>4<\/sub><sup>\u2212<\/sup><\/p>\n<\/li>\n<\/ol>\n<h1>Answers<\/h1>\n<ol>\n<li>A Br\u00f8nsted-Lowry acid is a proton donor. It does not necessarily increase the H<sup>+<\/sup> concentration in water.<\/li>\n<\/ol>\n<ol start=\"3\">\n<li>HBr +\u00a0H<sub>2<\/sub>O \u2192\u00a0H<sub>3<\/sub>O<sup>+<\/sup> +\u00a0Br<sup>\u2212<\/sup>; PD: HBr; PA: H<sub>2<\/sub>O<\/li>\n<\/ol>\n<ol start=\"5\">\n<li>C<sub>5<\/sub>H<sub>5<\/sub>N +\u00a0H<sub>2<\/sub>O \u2192\u00a0C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup> +\u00a0OH<sup>\u2212<\/sup>; PD: H<sub>2<\/sub>O; PA: C<sub>5<\/sub>H<sub>5<\/sub>N<\/li>\n<\/ol>\n<ol start=\"7\">\n<li>BL acid: H<sub>3<\/sub>PO<sub>4<\/sub>; BL base: OH<sup>\u2212<\/sup><\/li>\n<\/ol>\n<ol start=\"9\">\n<li>C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup> and C<sub>5<\/sub>H<sub>5<\/sub>NH<sup>+<\/sup><\/li>\n<\/ol>\n<ol start=\"11\">\n<li>H<sub>3<\/sub>O<sup>+<\/sup>; NH<sub>4<\/sub><sup>+<\/sup><\/li>\n<\/ol>\n<ol start=\"13\">\n<li>SO<sub>4<\/sub><sup>2\u2212<\/sup>; OH<sup>\u2212<\/sup><\/li>\n<\/ol>\n<ol start=\"15\">\n<li>HSO<sub>4<\/sub><sup>\u2212<\/sup> and SO<sub>4<\/sub><sup>2\u2212<\/sup>; PO<sub>4<\/sub><sup>3\u2212<\/sup> and HPO<sub>4<\/sub><sup>2\u2212<\/sup><\/li>\n<\/ol>\n<ol start=\"17\">\n<li>NH<sub>3<\/sub> and NH<sub>2<\/sub><sup>\u2212<\/sup>; C<sub>6<\/sub>H<sub>5<\/sub>O<sup>\u2212<\/sup> and C<sub>6<\/sub>H<sub>5<\/sub>OH<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":90,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7629","chapter","type-chapter","status-publish","hentry"],"part":7615,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7629","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":10,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7629\/revisions"}],"predecessor-version":[{"id":8933,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7629\/revisions\/8933"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7615"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7629\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7629"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7629"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7629"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7629"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}