{"id":595,"date":"2017-12-14T21:41:55","date_gmt":"2017-12-14T21:41:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/acid-base-titrations\/"},"modified":"2017-12-14T21:41:55","modified_gmt":"2017-12-14T21:41:55","slug":"acid-base-titrations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/acid-base-titrations\/","title":{"raw":"Acid-Base Titrations","rendered":"Acid-Base Titrations"},"content":{"raw":"<div class=\"section\" id=\"ball-ch12_s03\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch12_s03_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch12_s03_l01\"><li>Describe a titration experiment.<\/li>\n \t<li>Explain what an indicator does.<\/li>\n \t<li>Perform a titration calculation correctly.<\/li>\n<\/ol><\/div>\n<\/div>\n<p id=\"ball-ch12_s03_p01\" class=\"para editable block\">The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a <span class=\"margin_term\"><a class=\"glossterm\">titration<\/a><\/span>. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions.<\/p>\n<p id=\"ball-ch12_s03_p02\" class=\"para editable block\">In a titration, one reagent has a known concentration or amount, while the other reagent has an unknown concentration or amount. Typically, the known reagent (the <span class=\"margin_term\"><a class=\"glossterm\">titrant<\/a><\/span>) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the <span class=\"margin_term\"><a class=\"glossterm\">analyte<\/a><\/span>) may or may not be dissolved in solution (but usually is). The titrant is added to the analyte using a precisely calibrated volumetric delivery tube called a burette (also spelled buret; see <a class=\"xref\" href=\"#ball-ch12_s03_f01\">Figure 12.1 \"Equipment for Titrations\"<\/a>). The burette has markings to determine how much volume of solution has been added to the analyte. When the reaction is complete, it is said to be at the <span class=\"margin_term\"><a class=\"glossterm\">equivalence point<\/a><\/span>; the number of moles of titrant can be calculated from the concentration and the volume, and the balanced chemical equation can be used to determine the number of moles (and then concentration or mass) of the unknown reactant.<\/p>\n\n<div class=\"figure small editable block\" id=\"ball-ch12_s03_f01\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 12.1<\/span> Equipment for Titrations<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Titration-Equipment.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214144\/Titration-Equipment-1.png\" alt=\"Titration Equipment\" class=\"alignnone wp-image-4752\" width=\"400\" height=\"851\"\/><\/a><\/p>\n<p class=\"para\">A burette is a type of liquid dispensing system that can accurately indicate the volume of liquid dispensed.<\/p>\n\n<\/div>\n<p id=\"ball-ch12_s03_p03\" class=\"para editable block\">For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\"># mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl<\/span><\/span>\n<p id=\"ball-ch12_s03_p04\" class=\"para editable block\">We also have the balanced chemical reaction between HCl and NaOH:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">HCl +\u00a0NaOH \u2192\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p id=\"ball-ch12_s03_p05\" class=\"para editable block\">So we can construct a conversion factor to convert to number of moles of NaOH reacted:<\/p>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-23-at-10.49.25-AM.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214146\/Screen-Shot-2014-07-23-at-10.49.25-AM-1.png\" alt=\"Screen Shot 2014-07-23 at 10.49.25 AM\" class=\"aligncenter wp-image-3895\" width=\"429\" height=\"67\"\/><\/a>\n<p id=\"ball-ch12_s03_p06\" class=\"para editable block\">Then we convert this amount to mass, using the molar mass of NaOH (40.00 g\/mol):<\/p>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-23-at-10.49.29-AM.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214148\/Screen-Shot-2014-07-23-at-10.49.29-AM-1.png\" alt=\"Screen Shot 2014-07-23 at 10.49.29 AM\" class=\"aligncenter wp-image-3897\" width=\"424\" height=\"65\"\/><\/a>\n<p id=\"ball-ch12_s03_p07\" class=\"para editable block\">This is type of calculation is performed as part of a titration.<\/p>\n\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p id=\"ball-ch12_s03_p08\" class=\"para\">What mass of Ca(OH)<sub class=\"subscript\">2<\/sub> is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO<sub class=\"subscript\">3<\/sub>? The balanced chemical equation is as follows:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">2 HNO<sub class=\"subscript\">3<\/sub> +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s03_p09\" class=\"para\">In liters, the volume is 0.04402 L. We calculate the number of moles of titrant:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\"># moles HNO<sub class=\"subscript\">3<\/sub> = (0.04402 L)(0.0885 M) = 0.00390 mol HNO<sub class=\"subscript\">3<\/sub><\/span><\/span>\n<p id=\"ball-ch12_s03_p10\" class=\"para\">Using the balanced chemical equation, we can determine the number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> present in the analyte:<\/p>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/000390.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214151\/000390-1.png\" alt=\"000390\" class=\"alignnone wp-image-3898\" width=\"460\" height=\"75\"\/><\/a>\n<p id=\"ball-ch12_s03_p11\" class=\"para\">Then we convert this to a mass using the molar mass of Ca(OH)<sub class=\"subscript\">2<\/sub>:<\/p>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/000195.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214153\/000195-1.png\" alt=\"000195\" class=\"alignnone wp-image-3899\" width=\"438\" height=\"77\"\/><\/a>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s03_p12\" class=\"para\">What mass of H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02 NaOH \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s03_p13\" class=\"para\">0.182 g<\/p>\n\n<\/div>\n<p id=\"ball-ch12_s03_p14\" class=\"para editable block\">How does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an <span class=\"margin_term\"><a class=\"glossterm\">indicator<\/a><\/span>, a substance that changes color depending on the acidity or basicity of the solution. Because different indicators change colors at different levels of acidity, choosing the correct one is important in performing an accurate titration.<\/p>\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/qrcode.23437635.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214155\/qrcode.23437635-150x150-1.png\" alt=\"qrcode.23437635\" class=\"alignright size-thumbnail wp-image-3969\" width=\"150\" height=\"150\"\/><\/a>\n\nVideo source: Titration by keyj (<a href=\"https:\/\/viuvideos.viu.ca\/media\/Titration\/0_o7ixccav\">https:\/\/viuvideos.viu.ca\/media\/Titration\/0_o7ixccav<\/a>)\n<div class=\"key_takeaways editable block\" id=\"ball-ch12_s03_n03\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch12_s03_l02\" class=\"itemizedlist\"><li>A titration is the quantitative reaction of an acid and a base.<\/li>\n \t<li>Indicators are used to show that all the analyte has reacted with the titrant.<\/li>\n<\/ul><\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch12_s03_qs01\">\n<ol id=\"ball-ch12_s03_qs01_qd01\" class=\"qandadiv\"><li id=\"ball-ch12_s03_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p01\" class=\"para\">Define <em class=\"emphasis\">titration<\/em>.<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p03\" class=\"para\">What is the difference between the titrant and the analyte?<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p05\" class=\"para\">True or false: An acid is always the titrant. Explain your answer.<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p07\" class=\"para\">True or false: An analyte is always dissolved before reaction. Explain your answer.<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p09\" class=\"para\">If 55.60 mL of 0.2221 M HCl was needed to titrate a sample of NaOH to its equivalence point, what mass of NaOH was present?<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p11\" class=\"para\">If 16.33 mL of 0.6664 M KOH was needed to titrate a sample of HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> to its equivalence point, what mass of HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> was present?<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p13\" class=\"para\">It takes 45.66 mL of 0.1126 M HBr to titrate 25.00 mL of Ca(OH)<sub class=\"subscript\">2<\/sub> to its equivalence point. What is the original concentration of the Ca(OH)<sub class=\"subscript\">2<\/sub> solution?<\/p>\n\n<\/div><\/li>\n \t<li id=\"ball-ch12_s03_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p15\" class=\"para\">It takes 9.77 mL of 0.883 M H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> to titrate 15.00 mL of KOH to its equivalence point. What is the original concentration of the KOH solution?<\/p>\n\n<\/div><\/li>\n<\/ol><\/div>\n<b>Answers<\/b>\n\n<strong>1.<\/strong>\n\na chemical reaction performed in a quantitative fashion\n\n<strong>3.<\/strong>\n\nFalse; a base can be a titrant, or the reaction being performed may not even be an acid-base reaction.\n\n<strong>5.<\/strong>\n\n0.494 g\n\n<strong>7.<\/strong>\n\n0.1028 M\n\n<\/div>\n<\/div>\n<\/div>","rendered":"<div class=\"section\" id=\"ball-ch12_s03\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch12_s03_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch12_s03_l01\">\n<li>Describe a titration experiment.<\/li>\n<li>Explain what an indicator does.<\/li>\n<li>Perform a titration calculation correctly.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"ball-ch12_s03_p01\" class=\"para editable block\">The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a <span class=\"margin_term\"><a class=\"glossterm\">titration<\/a><\/span>. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions.<\/p>\n<p id=\"ball-ch12_s03_p02\" class=\"para editable block\">In a titration, one reagent has a known concentration or amount, while the other reagent has an unknown concentration or amount. Typically, the known reagent (the <span class=\"margin_term\"><a class=\"glossterm\">titrant<\/a><\/span>) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the <span class=\"margin_term\"><a class=\"glossterm\">analyte<\/a><\/span>) may or may not be dissolved in solution (but usually is). The titrant is added to the analyte using a precisely calibrated volumetric delivery tube called a burette (also spelled buret; see <a class=\"xref\" href=\"#ball-ch12_s03_f01\">Figure 12.1 &#8220;Equipment for Titrations&#8221;<\/a>). The burette has markings to determine how much volume of solution has been added to the analyte. When the reaction is complete, it is said to be at the <span class=\"margin_term\"><a class=\"glossterm\">equivalence point<\/a><\/span>; the number of moles of titrant can be calculated from the concentration and the volume, and the balanced chemical equation can be used to determine the number of moles (and then concentration or mass) of the unknown reactant.<\/p>\n<div class=\"figure small editable block\" id=\"ball-ch12_s03_f01\">\n<p class=\"title\"><span class=\"title-prefix\">Figure 12.1<\/span> Equipment for Titrations<\/p>\n<p class=\"title\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Titration-Equipment.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214144\/Titration-Equipment-1.png\" alt=\"Titration Equipment\" class=\"alignnone wp-image-4752\" width=\"400\" height=\"851\" \/><\/a><\/p>\n<p class=\"para\">A burette is a type of liquid dispensing system that can accurately indicate the volume of liquid dispensed.<\/p>\n<\/div>\n<p id=\"ball-ch12_s03_p03\" class=\"para editable block\">For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\"># mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl<\/span><\/span><\/p>\n<p id=\"ball-ch12_s03_p04\" class=\"para editable block\">We also have the balanced chemical reaction between HCl and NaOH:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">HCl +\u00a0NaOH \u2192\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch12_s03_p05\" class=\"para editable block\">So we can construct a conversion factor to convert to number of moles of NaOH reacted:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-23-at-10.49.25-AM.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214146\/Screen-Shot-2014-07-23-at-10.49.25-AM-1.png\" alt=\"Screen Shot 2014-07-23 at 10.49.25 AM\" class=\"aligncenter wp-image-3895\" width=\"429\" height=\"67\" \/><\/a><\/p>\n<p id=\"ball-ch12_s03_p06\" class=\"para editable block\">Then we convert this amount to mass, using the molar mass of NaOH (40.00 g\/mol):<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-23-at-10.49.29-AM.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214148\/Screen-Shot-2014-07-23-at-10.49.29-AM-1.png\" alt=\"Screen Shot 2014-07-23 at 10.49.29 AM\" class=\"aligncenter wp-image-3897\" width=\"424\" height=\"65\" \/><\/a><\/p>\n<p id=\"ball-ch12_s03_p07\" class=\"para editable block\">This is type of calculation is performed as part of a titration.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p id=\"ball-ch12_s03_p08\" class=\"para\">What mass of Ca(OH)<sub class=\"subscript\">2<\/sub> is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO<sub class=\"subscript\">3<\/sub>? The balanced chemical equation is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 HNO<sub class=\"subscript\">3<\/sub> +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s03_p09\" class=\"para\">In liters, the volume is 0.04402 L. We calculate the number of moles of titrant:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\"># moles HNO<sub class=\"subscript\">3<\/sub> = (0.04402 L)(0.0885 M) = 0.00390 mol HNO<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p id=\"ball-ch12_s03_p10\" class=\"para\">Using the balanced chemical equation, we can determine the number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> present in the analyte:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/000390.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214151\/000390-1.png\" alt=\"000390\" class=\"alignnone wp-image-3898\" width=\"460\" height=\"75\" \/><\/a><\/p>\n<p id=\"ball-ch12_s03_p11\" class=\"para\">Then we convert this to a mass using the molar mass of Ca(OH)<sub class=\"subscript\">2<\/sub>:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/000195.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214153\/000195-1.png\" alt=\"000195\" class=\"alignnone wp-image-3899\" width=\"438\" height=\"77\" \/><\/a><\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s03_p12\" class=\"para\">What mass of H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02 NaOH \u2192\u00a0Na<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s03_p13\" class=\"para\">0.182 g<\/p>\n<\/div>\n<p id=\"ball-ch12_s03_p14\" class=\"para editable block\">How does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an <span class=\"margin_term\"><a class=\"glossterm\">indicator<\/a><\/span>, a substance that changes color depending on the acidity or basicity of the solution. Because different indicators change colors at different levels of acidity, choosing the correct one is important in performing an accurate titration.<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/qrcode.23437635.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214155\/qrcode.23437635-150x150-1.png\" alt=\"qrcode.23437635\" class=\"alignright size-thumbnail wp-image-3969\" width=\"150\" height=\"150\" \/><\/a><\/p>\n<p>Video source: Titration by keyj (<a href=\"https:\/\/viuvideos.viu.ca\/media\/Titration\/0_o7ixccav\">https:\/\/viuvideos.viu.ca\/media\/Titration\/0_o7ixccav<\/a>)<\/p>\n<div class=\"key_takeaways editable block\" id=\"ball-ch12_s03_n03\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch12_s03_l02\" class=\"itemizedlist\">\n<li>A titration is the quantitative reaction of an acid and a base.<\/li>\n<li>Indicators are used to show that all the analyte has reacted with the titrant.<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch12_s03_qs01\">\n<ol id=\"ball-ch12_s03_qs01_qd01\" class=\"qandadiv\">\n<li id=\"ball-ch12_s03_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p01\" class=\"para\">Define <em class=\"emphasis\">titration<\/em>.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p03\" class=\"para\">What is the difference between the titrant and the analyte?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p05\" class=\"para\">True or false: An acid is always the titrant. Explain your answer.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p07\" class=\"para\">True or false: An analyte is always dissolved before reaction. Explain your answer.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p09\" class=\"para\">If 55.60 mL of 0.2221 M HCl was needed to titrate a sample of NaOH to its equivalence point, what mass of NaOH was present?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p11\" class=\"para\">If 16.33 mL of 0.6664 M KOH was needed to titrate a sample of HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> to its equivalence point, what mass of HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> was present?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p13\" class=\"para\">It takes 45.66 mL of 0.1126 M HBr to titrate 25.00 mL of Ca(OH)<sub class=\"subscript\">2<\/sub> to its equivalence point. What is the original concentration of the Ca(OH)<sub class=\"subscript\">2<\/sub> solution?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s03_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s03_qs01_p15\" class=\"para\">It takes 9.77 mL of 0.883 M H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> to titrate 15.00 mL of KOH to its equivalence point. What is the original concentration of the KOH solution?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p><b>Answers<\/b><\/p>\n<p><strong>1.<\/strong><\/p>\n<p>a chemical reaction performed in a quantitative fashion<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>False; a base can be a titrant, or the reaction being performed may not even be an acid-base reaction.<\/p>\n<p><strong>5.<\/strong><\/p>\n<p>0.494 g<\/p>\n<p><strong>7.<\/strong><\/p>\n<p>0.1028 M<\/p>\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-595\">\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>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/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":23485,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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