{"id":856,"date":"2018-03-20T16:18:15","date_gmt":"2018-03-20T16:18:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&#038;p=856"},"modified":"2018-08-08T16:38:23","modified_gmt":"2018-08-08T16:38:23","slug":"10-2-bronsted-lowry-definition-of-acids-and-bases","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/10-2-bronsted-lowry-definition-of-acids-and-bases\/","title":{"raw":"10.2 Br\u00f8nsted-Lowry Definition of Acids and Bases","rendered":"10.2 Br\u00f8nsted-Lowry Definition of Acids and Bases"},"content":{"raw":"<div id=\"navbar-top\" class=\"navbar\">\r\n<div class=\"navbar-part left\"><\/div>\r\n<\/div>\r\n<div id=\"book-content\">\r\n<div id=\"gob-ch10_s02\" class=\"section\" xml:lang=\"en\">\r\n<h2 class=\"title editable block\"><span class=\"title-prefix\">10.2<\/span> Br\u00f8nsted-Lowry Definition of Acids and Bases<\/h2>\r\n<div id=\"gob-ch10_s02_n01\" class=\"learning_objectives editable block\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3 class=\"title\">Learning Objectives<\/h3>\r\n<ol id=\"gob-ch10_s02_l01\" class=\"orderedlist\">\r\n \t<li>Recognize a compound as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/li>\r\n \t<li>Illustrate the proton transfer process that defines a Br\u00f8nsted-Lowry acid-base reaction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<span style=\"font-size: 1rem;text-align: initial\">Ammonia (NH<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1rem;text-align: initial\">) increases the hydroxide ion concentration in aqueous solution by reacting with water rather than releasing hydroxide ions directly. In fact, the Arrhenius definitions of an acid and a base focus on hydrogen ions and hydroxide ions. Are there more fundamental definitions for acids and bases?<\/span>\r\n\r\n<\/div>\r\n<p id=\"gob-ch10_s02_p02\" class=\"para editable block\">In 1923, the Danish scientist Johannes Br\u00f8nsted and the English scientist Thomas Lowry independently proposed new definitions for acids and bases. Rather than considering both hydrogen and hydroxide ions, they focused on the hydrogen ion only. A <span class=\"margin_term\"><span class=\"glossterm\">Br\u00f8nsted-Lowry acid<\/span><\/span>\u00a0is a compound that supplies a hydrogen ion in a reaction. A <span class=\"margin_term\"><span class=\"glossterm\">Br\u00f8nsted-Lowry base<\/span><\/span>, conversely, is a compound that accepts a hydrogen ion in a reaction. Thus, the Br\u00f8nsted-Lowry definitions of an acid and a base focus on the movement of hydrogen ions in a reaction, rather than on the production of hydrogen ions and hydroxide ions in an aqueous solution.<\/p>\r\n<p id=\"gob-ch10_s02_p03\" class=\"para editable block\">Let us use the reaction of ammonia in water to demonstrate the Br\u00f8nsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"gob-ch10_s02_p04\" class=\"para editable block\">What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:<\/p>\r\n\r\n<div class=\"informalfigure large block\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3101\/2018\/03\/20161714\/c9f5d4dfddd55fc337a56e77cb0154cd.jpg\" alt=\"image\" \/><\/div>\r\n<p id=\"gob-ch10_s02_p05\" class=\"para editable block\">Because the water molecule donates a hydrogen ion to the ammonia, it is the Br\u00f8nsted-Lowry acid, while the ammonia molecule\u2014which accepts the hydrogen ion\u2014is the Br\u00f8nsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Br\u00f8nsted-Lowry sense.<\/p>\r\n<p id=\"gob-ch10_s02_p06\" class=\"para editable block\">Is an Arrhenius acid like hydrochloric acid still an acid in the Br\u00f8nsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen <em class=\"emphasis\">atom<\/em> is a single proton surrounded by a single electron. To make the hydrogen <em class=\"emphasis\">ion<\/em>, we remove the electron, leaving a bare proton. Do we <em class=\"emphasis\">really<\/em> have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H<sup class=\"superscript\">+<\/sup> ion attaches itself to H<sub class=\"subscript\">2<\/sub>O to make H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>, which is called the <em class=\"emphasis\">hydronium ion<\/em>. For most purposes, H<sup class=\"superscript\">+<\/sup> and H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> represent the same species, but writing H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> instead of H<sup class=\"superscript\">+<\/sup> shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.<\/p>\r\n\r\n<div id=\"gob-ch10_s02_n02\" class=\"callout editable block\">\r\n<div class=\"textbox\">\r\n<h3 class=\"title\">Note<\/h3>\r\n<p id=\"gob-ch10_s02_p07\" class=\"para\">A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">+<\/sup> or H<sub class=\"subscript\">9<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> rather than H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>. It is simpler, however, to use H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"gob-ch10_s02_p08\" class=\"para editable block\">With this in mind, how do we define HCl as an acid in the Br\u00f8nsted-Lowry sense? Consider what happens when HCl is dissolved in H<sub class=\"subscript\">2<\/sub>O:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">HCl + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) + Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"gob-ch10_s02_p09\" class=\"para editable block\">We can depict this process using Lewis electron dot diagrams:<\/p>\r\n\r\n<div class=\"informalfigure large block\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3101\/2018\/03\/20161717\/1d3f9831c5173af6402646b6e19c2eef.jpg\" alt=\"image\" \/><\/div>\r\n<p id=\"gob-ch10_s02_p10\" class=\"para editable block\">Now we see that a hydrogen ion is transferred from the HCl molecule to the H<sub class=\"subscript\">2<\/sub>O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Br\u00f8nsted-Lowry acid; as a hydrogen ion acceptor, H<sub class=\"subscript\">2<\/sub>O is a Br\u00f8nsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Br\u00f8nsted-Lowry sense. Moreover, by the Br\u00f8nsted-Lowry definitions, H<sub class=\"subscript\">2<\/sub>O is a base in the formation of aqueous HCl. So the Br\u00f8nsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base\u2014although the Arrhenius definition would not have labeled H<sub class=\"subscript\">2<\/sub>O a base in this circumstance.<\/p>\r\n\r\n<div id=\"gob-ch10_s02_n03\" class=\"callout editable block\">\r\n<div class=\"textbox\">\r\n<h3 class=\"title\">Note<\/h3>\r\n<p id=\"gob-ch10_s02_p11\" class=\"para\">All Arrhenius acids and bases are Br\u00f8nsted-Lowry acids and bases as well. But not all Br\u00f8nsted-Lowry acids and bases are Arrhenius acids and bases.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_n04\" class=\"exercises block\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"gob-ch10_s02_p12\" class=\"para\">Aniline (C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub>) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n<p class=\"simpara\">Solution<\/p>\r\n<p id=\"gob-ch10_s02_p13\" class=\"para\">C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O are the reactants. When C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> accepts a proton from H<sub class=\"subscript\">2<\/sub>O, it gains an extra H and a positive charge and leaves an OH<sup class=\"superscript\">\u2212<\/sup> ion behind. The reaction is as follows:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"gob-ch10_s02_p14\" class=\"para\">Because C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> accepts a proton, it is the Br\u00f8nsted-Lowry base. The H<sub class=\"subscript\">2<\/sub>O molecule, because it donates a proton, is the Br\u00f8nsted-Lowry acid.<\/p>\r\n\r\n<\/div>\r\n<div id=\"gob-ch10_s02_qs01\" class=\"qandaset block\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\r\n<ol id=\"gob-ch10_s02_qs01_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s02_qs01_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs01_p01\" class=\"para\">Caffeine (C<sub class=\"subscript\">8<\/sub>H<sub class=\"subscript\">10<\/sub>N<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">2<\/sub>) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water molecule. Write the chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"gob-ch10_s02_p18\" class=\"para editable block\">The Br\u00f8nsted-Lowry definitions of an acid and a base can be applied to chemical reactions that occur in solvents other than water. The following example illustrates.<\/p>\r\n\r\n<div id=\"gob-ch10_s02_n06\" class=\"exercises block\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\n<p id=\"gob-ch10_s02_p19\" class=\"para\">Sodium amide (NaNH<sub class=\"subscript\">2<\/sub>) dissolves in methanol (CH<sub class=\"subscript\">3<\/sub>OH) and separates into sodium ions and amide ions (NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>). The amide ions react with methanol to make ammonia and the methoxide ion (CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup>). Write a balanced chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n<p class=\"simpara\">Solution<\/p>\r\n<p id=\"gob-ch10_s02_p20\" class=\"para\">The equation for the reaction is between NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> and CH<sub class=\"subscript\">3<\/sub>OH to make NH<sub class=\"subscript\">3<\/sub> and CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup> is as follows:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(solv) + CH<sub class=\"subscript\">3<\/sub>OH(\u2113) \u2192 NH<sub class=\"subscript\">3<\/sub>(solv) + CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup>(solv)<\/span><\/span>\r\n<p id=\"gob-ch10_s02_p21\" class=\"para\">The label <em class=\"emphasis\">(solv)<\/em> indicates that the species are dissolved in some solvent, in contrast to <em class=\"emphasis\">(aq)<\/em>, which specifies an aqueous (H<sub class=\"subscript\">2<\/sub>O) solution. In this reaction, we see that the NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ion accepts a proton from a CH<sub class=\"subscript\">3<\/sub>OH molecule to make an NH<sub class=\"subscript\">3<\/sub> molecule. Thus, as the proton acceptor, NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> is the Br\u00f8nsted-Lowry base. As the proton donor, CH<sub class=\"subscript\">3<\/sub>OH is the Br\u00f8nsted-Lowry acid.<\/p>\r\n\r\n<\/div>\r\n<div id=\"gob-ch10_s02_qs02\" class=\"qandaset block\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\r\n<ol id=\"gob-ch10_s02_qs02_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s02_qs02_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs02_p01\" class=\"para\">Pyridinium chloride (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHCl) dissolves in ethanol (C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH) and separates into pyridinium ions (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sup class=\"superscript\">+<\/sup>) and chloride ions. The pyridinium ion can transfer a hydrogen ion to a solvent molecule. Write a balanced chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_n08\" class=\"callout block\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">To Your Health: Br\u00f8nsted-Lowry Acid-Base Reactions in Pharmaceuticals<\/h3>\r\n<p id=\"gob-ch10_s02_p24\" class=\"para\">There are many interesting applications of Br\u00f8nsted-Lowry acid-base reactions in the pharmaceutical industry. For example, drugs often need to be water soluble for maximum effectiveness. However, many complex organic compounds are not soluble or are only slightly soluble in water. Fortunately, those drugs that contain proton-accepting nitrogen atoms (and there are a lot of them) can be reacted with dilute hydrochloric acid [HCl(aq)]. The nitrogen atoms\u2014acting as Br\u00f8nsted-Lowry bases\u2014accept the hydrogen ions from the acid to make an ion, which is usually much more soluble in water. The modified drug molecules can then be isolated as chloride salts:<\/p>\r\n<span class=\"informalequation\"> [latex]RN(sl\\,aq)\\,+\\,H^+(aq)\\,\u2192\\,RNH^+(aq)\\,_{---\\rightarrow}^{Cl^\u2212(aq)}\\,RNHCl(s)[\/latex]\u00a0 <\/span>\r\n<p id=\"gob-ch10_s02_p25\" class=\"para\">where RN represents some organic compound containing nitrogen. The label <em class=\"emphasis\">(sl aq)<\/em> means \u201cslightly aqueous,\u201d indicating that the compound RN is only slightly soluble. Drugs that are modified in this way are called <em class=\"emphasis\">hydrochloride salts<\/em>. Examples include the powerful painkiller codeine, which is commonly administered as codeine hydrochloride. Acids other than hydrochloric acid are also used. Hydrobromic acid, for example, gives <em class=\"emphasis\">hydrobromide salts<\/em>. Dextromethorphan, an ingredient in many cough medicines, is dispensed as dextromethorphan hydrobromide. The accompanying figure shows another medication as a hydrochloride salt.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_qs03\" class=\"qandaset block\">\r\n<div class=\"textbox exercises\">\r\n<h3 class=\"title\">Concept Review Exercise<\/h3>\r\n<ol id=\"gob-ch10_s02_qs03_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s02_qs03_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs03_p01\" class=\"para\">Give the definitions of a Br\u00f8nsted-Lowry acid and a Br\u00f8nsted-Lowry base.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Answer<\/h3>\r\n<div id=\"gob-ch10_s02_qs03_ans\" class=\"qandaset block\">\r\n<ol class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s02_qs03_qd01_qa01_ans\" class=\"qandaentry\">\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs03_p02_ans\" class=\"para\">A Br\u00f8nsted-Lowry acid is a proton donor, while a Br\u00f8nsted-Lowry base is a proton acceptor.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_n10\" class=\"key_takeaways editable block\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Takeaways<\/h3>\r\n<ul id=\"gob-ch10_s02_l02\" class=\"itemizedlist\">\r\n \t<li>A Br\u00f8nsted-Lowry acid is a proton donor, and a Br\u00f8nsted-Lowry base is a proton acceptor.<\/li>\r\n \t<li>Br\u00f8nsted-Lowry acid-base reactions are essentially proton transfer reactions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\r\n<div class=\"textbox exercises\">\r\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\r\n<h3 class=\"title\">Exercises<\/h3>\r\n<ol id=\"gob-ch10_s02_qs04_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p01\" class=\"para\">Label each reactant as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">HCl(aq) + NH<sub class=\"subscript\">3<\/sub>(aq) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p03\" class=\"para\">Label each reactant as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>O(\u2113) + N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>(aq) \u2192 N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa03\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p05\" class=\"para\">Explain why a Br\u00f8nsted-Lowry acid can be called a proton donor.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa04\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p07\" class=\"para\">Explain why a Br\u00f8nsted-Lowry base can be called a proton acceptor.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa05\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p09\" class=\"para\">Write the chemical equation of the reaction of ammonia in water and label the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa06\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p11\" class=\"para\">Write the chemical equation of the reaction of methylamine (CH<sub class=\"subscript\">3<\/sub>NH<sub class=\"subscript\">2<\/sub>) in water and label the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa07\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p13\" class=\"para\">Demonstrate that the dissolution of HNO<sub class=\"subscript\">3<\/sub> in water is actually a Br\u00f8nsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Br\u00f8nsted-Lowry acid and base.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa08\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p15\" class=\"para\">Identify the Br\u00f8nsted-Lowry acid and base in the following chemical equation:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">7<\/sub>NH<sub class=\"subscript\">2<\/sub>(aq) + H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) \u2192 C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">7<\/sub>NH<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">+<\/sup>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa09\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p17\" class=\"para\">Write the chemical equation for the reaction that occurs when cocaine hydrochloride (C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">22<\/sub>ClNO<sub class=\"subscript\">4<\/sub>) dissolves in water and donates a proton to a water molecule. (When hydrochlorides dissolve in water, they separate into chloride ions and the appropriate cation.)<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s02_qs04_qd01_qa10\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s02_qs04_p19\" class=\"para\">If codeine hydrobromide has the formula C<sub class=\"subscript\">18<\/sub>H<sub class=\"subscript\">22<\/sub>BrNO<sub class=\"subscript\">3<\/sub>, what is the formula of the parent compound codeine?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<div id=\"book-content\">\r\n<div id=\"gob-ch10_s02\" class=\"section\" xml:lang=\"en\">\r\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\r\n<h3 class=\"title\">Answers<\/h3>\r\n<\/div>\r\n<div id=\"gob-ch10_s02_qs04_ans\" class=\"qandaset block\">\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs04_p02_ans\" class=\"para\">1. HCl: Br\u00f8nsted-Lowry acid; NH<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry base<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs04_p06_ans\" class=\"para\">3. A Br\u00f8nsted-Lowry acid gives away an H<sup class=\"superscript\">+<\/sup> ion\u2014nominally, a proton\u2014in an acid-base reaction.<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs04_p10_ans\" class=\"para\">5. NH<sub class=\"subscript\">3<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> + OH<sup class=\"superscript\">\u2212<\/sup>; NH<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry base; H<sub class=\"subscript\">2<\/sub>O: Br\u00f8nsted-Lowry acid<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs04_p14_ans\" class=\"para\">7. HNO<sub class=\"subscript\">3<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> + NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>; HNO<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry acid; H<sub class=\"subscript\">2<\/sub>O: Br\u00f8nsted-Lowry base<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s02_qs04_p18_ans\" class=\"para\">9. C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">22<\/sub>NO<sub class=\"subscript\">4+<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> + C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">4<\/sub><\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"navbar-bottom\" class=\"navbar\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"navbar-top\" class=\"navbar\">\n<div class=\"navbar-part left\"><\/div>\n<\/div>\n<div id=\"book-content\">\n<div id=\"gob-ch10_s02\" class=\"section\" xml:lang=\"en\">\n<h2 class=\"title editable block\"><span class=\"title-prefix\">10.2<\/span> Br\u00f8nsted-Lowry Definition of Acids and Bases<\/h2>\n<div id=\"gob-ch10_s02_n01\" class=\"learning_objectives editable block\">\n<div class=\"textbox learning-objectives\">\n<h3 class=\"title\">Learning Objectives<\/h3>\n<ol id=\"gob-ch10_s02_l01\" class=\"orderedlist\">\n<li>Recognize a compound as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/li>\n<li>Illustrate the proton transfer process that defines a Br\u00f8nsted-Lowry acid-base reaction.<\/li>\n<\/ol>\n<\/div>\n<p><span style=\"font-size: 1rem;text-align: initial\">Ammonia (NH<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1rem;text-align: initial\">) increases the hydroxide ion concentration in aqueous solution by reacting with water rather than releasing hydroxide ions directly. In fact, the Arrhenius definitions of an acid and a base focus on hydrogen ions and hydroxide ions. Are there more fundamental definitions for acids and bases?<\/span><\/p>\n<\/div>\n<p id=\"gob-ch10_s02_p02\" class=\"para editable block\">In 1923, the Danish scientist Johannes Br\u00f8nsted and the English scientist Thomas Lowry independently proposed new definitions for acids and bases. Rather than considering both hydrogen and hydroxide ions, they focused on the hydrogen ion only. A <span class=\"margin_term\"><span class=\"glossterm\">Br\u00f8nsted-Lowry acid<\/span><\/span>\u00a0is a compound that supplies a hydrogen ion in a reaction. A <span class=\"margin_term\"><span class=\"glossterm\">Br\u00f8nsted-Lowry base<\/span><\/span>, conversely, is a compound that accepts a hydrogen ion in a reaction. Thus, the Br\u00f8nsted-Lowry definitions of an acid and a base focus on the movement of hydrogen ions in a reaction, rather than on the production of hydrogen ions and hydroxide ions in an aqueous solution.<\/p>\n<p id=\"gob-ch10_s02_p03\" class=\"para editable block\">Let us use the reaction of ammonia in water to demonstrate the Br\u00f8nsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s02_p04\" class=\"para editable block\">What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:<\/p>\n<div class=\"informalfigure large block\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3101\/2018\/03\/20161714\/c9f5d4dfddd55fc337a56e77cb0154cd.jpg\" alt=\"image\" \/><\/div>\n<p id=\"gob-ch10_s02_p05\" class=\"para editable block\">Because the water molecule donates a hydrogen ion to the ammonia, it is the Br\u00f8nsted-Lowry acid, while the ammonia molecule\u2014which accepts the hydrogen ion\u2014is the Br\u00f8nsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Br\u00f8nsted-Lowry sense.<\/p>\n<p id=\"gob-ch10_s02_p06\" class=\"para editable block\">Is an Arrhenius acid like hydrochloric acid still an acid in the Br\u00f8nsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen <em class=\"emphasis\">atom<\/em> is a single proton surrounded by a single electron. To make the hydrogen <em class=\"emphasis\">ion<\/em>, we remove the electron, leaving a bare proton. Do we <em class=\"emphasis\">really<\/em> have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H<sup class=\"superscript\">+<\/sup> ion attaches itself to H<sub class=\"subscript\">2<\/sub>O to make H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>, which is called the <em class=\"emphasis\">hydronium ion<\/em>. For most purposes, H<sup class=\"superscript\">+<\/sup> and H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> represent the same species, but writing H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> instead of H<sup class=\"superscript\">+<\/sup> shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.<\/p>\n<div id=\"gob-ch10_s02_n02\" class=\"callout editable block\">\n<div class=\"textbox\">\n<h3 class=\"title\">Note<\/h3>\n<p id=\"gob-ch10_s02_p07\" class=\"para\">A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">+<\/sup> or H<sub class=\"subscript\">9<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> rather than H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>. It is simpler, however, to use H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>.<\/p>\n<\/div>\n<\/div>\n<p id=\"gob-ch10_s02_p08\" class=\"para editable block\">With this in mind, how do we define HCl as an acid in the Br\u00f8nsted-Lowry sense? Consider what happens when HCl is dissolved in H<sub class=\"subscript\">2<\/sub>O:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">HCl + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) + Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s02_p09\" class=\"para editable block\">We can depict this process using Lewis electron dot diagrams:<\/p>\n<div class=\"informalfigure large block\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3101\/2018\/03\/20161717\/1d3f9831c5173af6402646b6e19c2eef.jpg\" alt=\"image\" \/><\/div>\n<p id=\"gob-ch10_s02_p10\" class=\"para editable block\">Now we see that a hydrogen ion is transferred from the HCl molecule to the H<sub class=\"subscript\">2<\/sub>O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Br\u00f8nsted-Lowry acid; as a hydrogen ion acceptor, H<sub class=\"subscript\">2<\/sub>O is a Br\u00f8nsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Br\u00f8nsted-Lowry sense. Moreover, by the Br\u00f8nsted-Lowry definitions, H<sub class=\"subscript\">2<\/sub>O is a base in the formation of aqueous HCl. So the Br\u00f8nsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base\u2014although the Arrhenius definition would not have labeled H<sub class=\"subscript\">2<\/sub>O a base in this circumstance.<\/p>\n<div id=\"gob-ch10_s02_n03\" class=\"callout editable block\">\n<div class=\"textbox\">\n<h3 class=\"title\">Note<\/h3>\n<p id=\"gob-ch10_s02_p11\" class=\"para\">All Arrhenius acids and bases are Br\u00f8nsted-Lowry acids and bases as well. But not all Br\u00f8nsted-Lowry acids and bases are Arrhenius acids and bases.<\/p>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s02_n04\" class=\"exercises block\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"gob-ch10_s02_p12\" class=\"para\">Aniline (C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub>) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"gob-ch10_s02_p13\" class=\"para\">C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O are the reactants. When C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> accepts a proton from H<sub class=\"subscript\">2<\/sub>O, it gains an extra H and a positive charge and leaves an OH<sup class=\"superscript\">\u2212<\/sup> ion behind. The reaction is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s02_p14\" class=\"para\">Because C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">2<\/sub> accepts a proton, it is the Br\u00f8nsted-Lowry base. The H<sub class=\"subscript\">2<\/sub>O molecule, because it donates a proton, is the Br\u00f8nsted-Lowry acid.<\/p>\n<\/div>\n<div id=\"gob-ch10_s02_qs01\" class=\"qandaset block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\n<ol id=\"gob-ch10_s02_qs01_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s02_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs01_p01\" class=\"para\">Caffeine (C<sub class=\"subscript\">8<\/sub>H<sub class=\"subscript\">10<\/sub>N<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">2<\/sub>) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water molecule. Write the chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"gob-ch10_s02_p18\" class=\"para editable block\">The Br\u00f8nsted-Lowry definitions of an acid and a base can be applied to chemical reactions that occur in solvents other than water. The following example illustrates.<\/p>\n<div id=\"gob-ch10_s02_n06\" class=\"exercises block\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"gob-ch10_s02_p19\" class=\"para\">Sodium amide (NaNH<sub class=\"subscript\">2<\/sub>) dissolves in methanol (CH<sub class=\"subscript\">3<\/sub>OH) and separates into sodium ions and amide ions (NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>). The amide ions react with methanol to make ammonia and the methoxide ion (CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup>). Write a balanced chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"gob-ch10_s02_p20\" class=\"para\">The equation for the reaction is between NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> and CH<sub class=\"subscript\">3<\/sub>OH to make NH<sub class=\"subscript\">3<\/sub> and CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup> is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(solv) + CH<sub class=\"subscript\">3<\/sub>OH(\u2113) \u2192 NH<sub class=\"subscript\">3<\/sub>(solv) + CH<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">\u2212<\/sup>(solv)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s02_p21\" class=\"para\">The label <em class=\"emphasis\">(solv)<\/em> indicates that the species are dissolved in some solvent, in contrast to <em class=\"emphasis\">(aq)<\/em>, which specifies an aqueous (H<sub class=\"subscript\">2<\/sub>O) solution. In this reaction, we see that the NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ion accepts a proton from a CH<sub class=\"subscript\">3<\/sub>OH molecule to make an NH<sub class=\"subscript\">3<\/sub> molecule. Thus, as the proton acceptor, NH<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> is the Br\u00f8nsted-Lowry base. As the proton donor, CH<sub class=\"subscript\">3<\/sub>OH is the Br\u00f8nsted-Lowry acid.<\/p>\n<\/div>\n<div id=\"gob-ch10_s02_qs02\" class=\"qandaset block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\n<ol id=\"gob-ch10_s02_qs02_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s02_qs02_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs02_p01\" class=\"para\">Pyridinium chloride (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHCl) dissolves in ethanol (C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH) and separates into pyridinium ions (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sup class=\"superscript\">+<\/sup>) and chloride ions. The pyridinium ion can transfer a hydrogen ion to a solvent molecule. Write a balanced chemical equation for this process and identify the Br\u00f8nsted-Lowry acid and base.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s02_n08\" class=\"callout block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">To Your Health: Br\u00f8nsted-Lowry Acid-Base Reactions in Pharmaceuticals<\/h3>\n<p id=\"gob-ch10_s02_p24\" class=\"para\">There are many interesting applications of Br\u00f8nsted-Lowry acid-base reactions in the pharmaceutical industry. For example, drugs often need to be water soluble for maximum effectiveness. However, many complex organic compounds are not soluble or are only slightly soluble in water. Fortunately, those drugs that contain proton-accepting nitrogen atoms (and there are a lot of them) can be reacted with dilute hydrochloric acid [HCl(aq)]. The nitrogen atoms\u2014acting as Br\u00f8nsted-Lowry bases\u2014accept the hydrogen ions from the acid to make an ion, which is usually much more soluble in water. The modified drug molecules can then be isolated as chloride salts:<\/p>\n<p><span class=\"informalequation\"> [latex]RN(sl\\,aq)\\,+\\,H^+(aq)\\,\u2192\\,RNH^+(aq)\\,_{---\\rightarrow}^{Cl^\u2212(aq)}\\,RNHCl(s)[\/latex]\u00a0 <\/span><\/p>\n<p id=\"gob-ch10_s02_p25\" class=\"para\">where RN represents some organic compound containing nitrogen. The label <em class=\"emphasis\">(sl aq)<\/em> means \u201cslightly aqueous,\u201d indicating that the compound RN is only slightly soluble. Drugs that are modified in this way are called <em class=\"emphasis\">hydrochloride salts<\/em>. Examples include the powerful painkiller codeine, which is commonly administered as codeine hydrochloride. Acids other than hydrochloric acid are also used. Hydrobromic acid, for example, gives <em class=\"emphasis\">hydrobromide salts<\/em>. Dextromethorphan, an ingredient in many cough medicines, is dispensed as dextromethorphan hydrobromide. The accompanying figure shows another medication as a hydrochloride salt.<\/p>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s02_qs03\" class=\"qandaset block\">\n<div class=\"textbox exercises\">\n<h3 class=\"title\">Concept Review Exercise<\/h3>\n<ol id=\"gob-ch10_s02_qs03_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s02_qs03_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs03_p01\" class=\"para\">Give the definitions of a Br\u00f8nsted-Lowry acid and a Br\u00f8nsted-Lowry base.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Answer<\/h3>\n<div id=\"gob-ch10_s02_qs03_ans\" class=\"qandaset block\">\n<ol class=\"qandadiv\">\n<li id=\"gob-ch10_s02_qs03_qd01_qa01_ans\" class=\"qandaentry\">\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs03_p02_ans\" class=\"para\">A Br\u00f8nsted-Lowry acid is a proton donor, while a Br\u00f8nsted-Lowry base is a proton acceptor.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s02_n10\" class=\"key_takeaways editable block\">\n<div class=\"textbox key-takeaways\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"gob-ch10_s02_l02\" class=\"itemizedlist\">\n<li>A Br\u00f8nsted-Lowry acid is a proton donor, and a Br\u00f8nsted-Lowry base is a proton acceptor.<\/li>\n<li>Br\u00f8nsted-Lowry acid-base reactions are essentially proton transfer reactions.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\n<div class=\"textbox exercises\">\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\n<h3 class=\"title\">Exercises<\/h3>\n<ol id=\"gob-ch10_s02_qs04_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s02_qs04_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p01\" class=\"para\">Label each reactant as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">HCl(aq) + NH<sub class=\"subscript\">3<\/sub>(aq) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p03\" class=\"para\">Label each reactant as a Br\u00f8nsted-Lowry acid or a Br\u00f8nsted-Lowry base.<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>O(\u2113) + N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">4<\/sub>(aq) \u2192 N<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p05\" class=\"para\">Explain why a Br\u00f8nsted-Lowry acid can be called a proton donor.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p07\" class=\"para\">Explain why a Br\u00f8nsted-Lowry base can be called a proton acceptor.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p09\" class=\"para\">Write the chemical equation of the reaction of ammonia in water and label the Br\u00f8nsted-Lowry acid and base.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p11\" class=\"para\">Write the chemical equation of the reaction of methylamine (CH<sub class=\"subscript\">3<\/sub>NH<sub class=\"subscript\">2<\/sub>) in water and label the Br\u00f8nsted-Lowry acid and base.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p13\" class=\"para\">Demonstrate that the dissolution of HNO<sub class=\"subscript\">3<\/sub> in water is actually a Br\u00f8nsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Br\u00f8nsted-Lowry acid and base.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p15\" class=\"para\">Identify the Br\u00f8nsted-Lowry acid and base in the following chemical equation:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">7<\/sub>NH<sub class=\"subscript\">2<\/sub>(aq) + H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) \u2192 C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">7<\/sub>NH<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">+<\/sup>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p17\" class=\"para\">Write the chemical equation for the reaction that occurs when cocaine hydrochloride (C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">22<\/sub>ClNO<sub class=\"subscript\">4<\/sub>) dissolves in water and donates a proton to a water molecule. (When hydrochlorides dissolve in water, they separate into chloride ions and the appropriate cation.)<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s02_qs04_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s02_qs04_p19\" class=\"para\">If codeine hydrobromide has the formula C<sub class=\"subscript\">18<\/sub>H<sub class=\"subscript\">22<\/sub>BrNO<sub class=\"subscript\">3<\/sub>, what is the formula of the parent compound codeine?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<div id=\"book-content\">\n<div id=\"gob-ch10_s02\" class=\"section\" xml:lang=\"en\">\n<div id=\"gob-ch10_s02_qs04\" class=\"qandaset block\">\n<h3 class=\"title\">Answers<\/h3>\n<\/div>\n<div id=\"gob-ch10_s02_qs04_ans\" class=\"qandaset block\">\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs04_p02_ans\" class=\"para\">1. HCl: Br\u00f8nsted-Lowry acid; NH<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry base<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs04_p06_ans\" class=\"para\">3. A Br\u00f8nsted-Lowry acid gives away an H<sup class=\"superscript\">+<\/sup> ion\u2014nominally, a proton\u2014in an acid-base reaction.<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs04_p10_ans\" class=\"para\">5. NH<sub class=\"subscript\">3<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> + OH<sup class=\"superscript\">\u2212<\/sup>; NH<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry base; H<sub class=\"subscript\">2<\/sub>O: Br\u00f8nsted-Lowry acid<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs04_p14_ans\" class=\"para\">7. HNO<sub class=\"subscript\">3<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> + NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>; HNO<sub class=\"subscript\">3<\/sub>: Br\u00f8nsted-Lowry acid; H<sub class=\"subscript\">2<\/sub>O: Br\u00f8nsted-Lowry base<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s02_qs04_p18_ans\" class=\"para\">9. C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">22<\/sub>NO<sub class=\"subscript\">4+<\/sub> + H<sub class=\"subscript\">2<\/sub>O \u2192 H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> + C<sub class=\"subscript\">17<\/sub>H<sub class=\"subscript\">21<\/sub>NO<sub class=\"subscript\">4<\/sub><\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"navbar-bottom\" class=\"navbar\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\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-856\">\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>The Basics of General, Organic, and Biological Chemistry v. 1.0. <strong>Provided by<\/strong>: Saylor Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/\">https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em>. <strong>License Terms<\/strong>: This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work&#039;s original creator or licensor.<\/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":44985,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"The Basics of General, Organic, and Biological Chemistry v. 1.0\",\"author\":\"\",\"organization\":\"Saylor Academy\",\"url\":\"https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work\\'s original creator or licensor.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-856","chapter","type-chapter","status-publish","hentry"],"part":838,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/856","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/856\/revisions"}],"predecessor-version":[{"id":3130,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/856\/revisions\/3130"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/parts\/838"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/856\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/media?parent=856"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=856"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/contributor?post=856"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/license?post=856"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}