{"id":2823,"date":"2016-08-24T19:39:23","date_gmt":"2016-08-24T19:39:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2823"},"modified":"2016-08-24T22:44:04","modified_gmt":"2016-08-24T22:44:04","slug":"titration","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/titration\/","title":{"raw":"Titration","rendered":"Titration"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define equivalence point.<\/li>\r\n \t<li>Describe how to perform a titration experiment.<\/li>\r\n \t<li>Perform calculations to determine concentration of unknown acid or base.<\/li>\r\n \t<li>Describe titration curves of acid-base neutralization reactions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"x-ck12-VGl0cmF0aW9uIEV4cGVyaW1lbnQ.\">Titration Experiment<\/h2>\r\n<div class=\"textbox examples\">\r\n<h3><strong>Didn\u2019t that used to be French fries? <\/strong><\/h3>\r\n<p id=\"x-ck12-YjhiYmY3YzQ0ZDUxMWY3NjA4NzI0MzY4YTM4NjQxYjU.-tik\">A lot of research is going on these days involving the development of biodiesel fuels. Often this material can be made from used vegetable oils. The vegetable oil is treated with lye to create the biofuel. In the oils is a variable amount of acid that needs to be determined so the workers will know how much lye to add to make the final fuel. Before the lye is added, the native vegetable oil is titrated to find out how much free acid is present. Then the amount of lye added can be adjusted to take into account the amount needed to neutralize these free acids.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213332\/20140811155916481045.jpeg\" alt=\"Biodiesel synthesis requires the amount of acid to be determined before adding lye\" width=\"250\" \/>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YjNmZmVjMDMyNDgyZDA0ZDQxNjRjYWNkOGU1ODlkZDA.-owf\">In the neutralization of hydrochloric acid by sodium hydroxide, the mole ratio of acid to base is 1:1.<\/p>\r\n<p id=\"x-ck12-qw6\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211839\/4b5dd661a2a365a9ee36f5fc7737474b.png\" alt=\"text{HCl}(aq)+text{NaOH}(aq) rightarrow text{NaCl}(aq)+text{H}_2text{O}(l)\" width=\"343\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-M2Q1Zjk0MDdmY2FkYjM1Mzg2NGM0NDA0MTVkZjgyZWI.-tec\">One mole of HCl would be fully neutralized by one mole of NaOH. If instead the hydrochloric acid was reacted with barium hydroxide, the mole ratio would be 2:1.<\/p>\r\n<p id=\"x-ck12-0bt\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213334\/63259ec0f75dffbad96f90f8350cf63e.png\" alt=\"2text{HCl}(aq)+text{Ba}(text{OH})_2(aq) rightarrow text{BaCl}_2(aq)+2text{H}_2text{O}(l)\" width=\"387\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-MDRhNGI4MjljZDA5NzRiMDZkODEzZTBhYTUwMTVkMDA.-sda\">Now two moles of HCl would be required to neutralize one mole of Ba(OH) <sub> 2 <\/sub> . The mole ratio insures that the number of moles of H <sup> + <\/sup> ions supplied by the acid is equal to the number of OH <sup> \u2212 <\/sup> ions supplied by the base. This must be the case for neutralization to occur. The <strong> equivalence point <\/strong> is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.<\/p>\r\n<p id=\"x-ck12-ZDM5ZWM0MGFjN2YzOWI4N2I2NmQ1MmY1YTFkZTBjN2Q.-u6p\">In the laboratory, it is useful to have an experiment where the unknown concentration of an acid or a base can be determined. This can be accomplished by performing a controlled neutralization reaction. A <strong> titration <\/strong> is an experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. Many titrations are acid-base neutralization reactions, though other types of titrations can also be performed.<\/p>\r\n<p id=\"x-ck12-NmIxNzAxMDYxNzRjM2EzZDdiNjZmMDY4NTI1MjJkOWM.-xan\">In order to perform an acid-base titration, the chemist must have a way to visually detect that the neutralization reaction has occurred. An <strong> indicator <\/strong> is a substance that has a distinctly different color when in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein. Solutions in which a few drops of phenolphthalein have been added turn from colorless to brilliant pink as the solution turns from acidic to basic. The steps in a titration reaction are outlined below.<\/p>\r\n\r\n<ol id=\"x-ck12-YzdiMThlZjIyZTlkNTY2MjU1N2Y1ZDkzY2E4ZjYxNTI.-npb\">\r\n \t<li>A measured volume of an acid of unknown concentration is added to an Erlenmeyer flask.<\/li>\r\n \t<li>Several drops of an indicator are added to the acid and mixed by swirling the flask.<\/li>\r\n \t<li>A buret is filled with the base solution of known molarity.<\/li>\r\n \t<li>The stopcock of the buret is opened and base is slowly added to the acid while the flask is constantly swirled to insure mixing. The stopcock is closed at the exact point at which the indicator just changes color.<\/li>\r\n<\/ol>\r\n<div id=\"x-ck12-YWE5OThkMjA3NjllNDBiNjBhZDAzNzYyZWM3ZDc5Y2M.-xpv\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"192\"]<img id=\"x-ck12-OTgwNDUtMTM2NDk4MjMwMi00MS0yMS01Ng..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213335\/20140811155916680904.jpeg\" alt=\"Phenolphthalein turns pink in basic solutions\" width=\"192\" height=\"172\" longdesc=\"Phenolphthalein%20in%20basic%20solution.\" \/> Figure 1.\u00a0Phenolphthalein in basic solution.[\/caption]\r\n\r\nThe <strong> standard solution <\/strong> is the solution in a titration whose concentration is known. In the titration described above the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly so that the point at which the indicator changes color can be found accurately.\r\n\r\nThe <strong> end point <\/strong> of a titration is the point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by a faint pink color.\r\n\r\n<\/div>\r\n<h2 id=\"x-ck12-VGl0cmF0aW9uIENhbGN1bGF0aW9ucw..\">Titration Calculations<\/h2>\r\n<div class=\"textbox examples\">\r\n<h3>How is soap made?<\/h3>\r\n<p id=\"x-ck12-NGIwNDYzMmJjOGMyY2IxYWYxZGVhOWQ3NTY1YWUxNWQ.-hm3\"><span class=\"x-ck12-img-inline\"><img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213337\/20140811155916755515.jpeg\" alt=\"The calculation of saponification number in the production of soap is typically found through titration\" width=\"233\" height=\"175\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZWViZmZlM2Y5MTRiMjE2Nzc3NTg1YjlmMjQ3YWZmMmQ.-ris\">The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product.<\/p>\r\nThe fat is heated with a known amount of base (usually NaOH or KOH). After hydrolysis is complete, the left-over base is titrated to determine how much was needed to hydrolyze the fat sample.\r\n\r\n<\/div>\r\n<h3>Titration Calculations<\/h3>\r\n<p id=\"x-ck12-NGY1MTMyOGZmZmM1ZjI4ZjE2NTMwNThiOTJhMWM1MjQ.-alb\">At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.<\/p>\r\nmoles acid = moles base\r\n<p id=\"x-ck12-YzMwMTE1ZjBiNDY2OWFjNjhiMTQ3MjFjZjc0YWUxZDg.-zkw\">Recall that the molarity (<em>M<\/em>)\u00a0of a solution is defined as the moles of the solute divided by the liters of solution (<em>L<\/em>). So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.<\/p>\r\n<p id=\"x-ck12-jc8\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213341\/7617be94d4cc9da7450f0ec91ebdf9f4.png\" alt=\"text{moles solute} = M times L\" width=\"173\" height=\"14\" \/><\/p>\r\n<p id=\"x-ck12-ZjU1Yjk3NGY3ZmI5MWE1NmVlMDQwZTJhM2ZhNzRjZGE.-dlx\">We can then set the moles of acid equal to the moles of base.<\/p>\r\n<p id=\"x-ck12-qe5\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213342\/724089f5770bbe6377231cc383015a42.png\" alt=\"M_A times V_A=M_B times V_B\" width=\"165\" height=\"15\" \/><\/p>\r\n<p id=\"x-ck12-ODlhZDkxNTVmNTk5OTU3NjlhOWU3ZDc2ODhlNGE2NWM.-eqm\"><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213342\/f094c5600c0d02a8ec44fbd0b26935d0.png\" alt=\"M_A\" width=\"27\" height=\"15\" \/> \u00a0is the molarity of the acid, while\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/43f078a33f698a7025f295f338e19214.png\" alt=\"M_B\" width=\"27\" height=\"15\" \/> is the molarity of the base.\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/7badc172c5b50023ee2e75b381450c4d.png\" alt=\"V_A\" width=\"20\" height=\"15\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/f2d1aa5f5039db0bdf46158eb1cde675.png\" alt=\"V_B\" width=\"20\" height=\"15\" \/> are the volumes of the acid and base, respectively.<\/p>\r\n<p id=\"x-ck12-MDc4NTI2N2NjZThmNTJlZDg2NGVhMDI5ZWM1MjVlYzg.-akd\">Suppose that a titration is performed and 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated against 15.00 mL of HCl of unknown concentration. The above equation can be used to solve for the molarity of the acid.<\/p>\r\n<p id=\"x-ck12-j3q\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213344\/815ee7cbc6d01dad16fa9e7e3ee6821f.png\" alt=\"M_A=frac{M_B times V_B}{V_A}=frac{0.500 text{M} times 20.70 text{mL}}{15.00 text{mL}}=0.690 text{M}\" width=\"400\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-NDljOTk4Y2U1M2M1MTI4ZGQwOGM4YmE2NzQ0OWNmMjc.-jug\">The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.<\/p>\r\n<p id=\"x-ck12-MDdiZTNmNWNmMTEyNTBlMDkzMDVhNjNkY2ZhY2JhM2Y.-voc\">The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The sample problem below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h4>Sample Problem: Titration<\/h4>\r\n<p id=\"x-ck12-OGY1YmRkMmU3Mjg1MTdjNWYyOWEzN2M5OTFmMDNlYjE.-qbm\">In a titration of sulfuric acid against sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H<sub>2<\/sub>SO<sub>4<\/sub>. Calculate the molarity of the sulfuric acid.<\/p>\r\n<p id=\"x-ck12-OGM1ZTljNzdiMjc4MjQyMjY3OTljNDI4OTk2ZmY0MTE.-o8g\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-byj\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-OTA4MjA5ZmZlYzljZGNlZGZlNTZlZGI4MDNmM2JmNWM.-vuc\">\r\n \t<li>molarity NaOH = 0.250 M<\/li>\r\n \t<li>volume NaOH = 32.20 mL<\/li>\r\n \t<li>volume H <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 26.60 mL<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NzNiNGU0YWI4NDkzN2MxNTQyYTUzOTViMzRiMzY4ZTI.-rwf\"><span class=\"x-ck12-underline\"> Unkonwn <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YmExOGMzZGE4Y2JhNzVjYmVhM2Q4ZGRiNTg0YzZlNmU.-kk3\">\r\n \t<li>molarity H <sub> 2 <\/sub> SO <sub> 4 <\/sub> = ?<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZGRmMTAxMTkwNTZhMjBjZjI2MTE1Y2QwZDM5ODEwZjE.-kct\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213345\/377340a89ea67f537ffdfbbde8ed62ec.png\" alt=\"text{equation} qquad text{H}_2 text{SO}_4(aq)+2text{NaOH}(aq) rightarrow text{Na}_2text{SO}_4(aq)+2text{H}_2text{O}(l)\" width=\"504\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-YzM2MDJiMTY0ZDY2NzlmMDc2OTQzODlmOTdmZTlmOTk.-yli\">First determine the moles of NaOH in the reaction. From the mole ratio, calculate the moles of H <sub> 2 <\/sub> SO <sub> 4 <\/sub> that reacted. Finally, divide the moles H <sub> 2 <\/sub> SO <sub> 4 <\/sub> by its volume to get the molarity.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-mu2\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-qea\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213347\/75ecadf37aa4f4b55d37d61178ad6acb.png\" alt=\"&amp; text{mol NaOH}=M times L=0.250 text{M} times 0.03220 text{L}=8.05 times 10^{-3} text{mol} NaOH \\&amp; 8.05 times 10^{-3} text{mol NaOH} times frac{1 text{mol H}_2text{SO}_4}{2 text{mol NaOH}}=4.03 times 10^{-3} text{mol H}_2text{SO}_4 \\&amp; frac{4.03 times 10^{-3} text{mol H}_2text{SO}_4}{0.02660 text{L}}=0.151 text{M H}_2text{SO}_4\" width=\"558\" height=\"113\" \/><\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-mbl\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NzZkMDhiNzI4MTAzMmIxODc0NWNlNmVkY2Y2ODc5NDM.-vfg\">The volume of H<sub>2<\/sub>SO<sub>4 <\/sub> required is smaller than the volume of NaOH because of the two hydrogen ions contributed by each molecule.<\/p>\r\n\r\n<\/div>\r\n<h2 id=\"x-ck12-VGl0cmF0aW9uIEN1cnZlcw..\">Titration Curves<\/h2>\r\n<p id=\"x-ck12-MDMzNjgzOGVhYjQ0Yjc0MTkxMWZiYjQ1YmM3YmY1OGI.-ykm\"><\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3><strong>Where did graphs come from? <\/strong><\/h3>\r\n<p id=\"x-ck12-OTIyOWY2NDg5MmM2NjM2YmIzZTI1ZWVjNDhmYmRlMzk.-bcm\">The\u00a0 <em>x\u2013y<\/em>\u00a0plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596\u20131650). His studies in mathematics led him to develop what was known as \u201cCartesian geometry,\u201d including the idea of our current graphs. The coordinates are often referred to as Cartesian coordinates.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213349\/20140811155916885001.jpeg\" alt=\"Cartesian graphs are often used to represent the course of a titration\" width=\"200\" \/>\r\n\r\n<\/div>\r\n<h3>Titration Curves<\/h3>\r\n<p id=\"x-ck12-ZmIzYjk4ODcwZDI3MzQ5N2JhZmIwMzY0MDIyYTU3OTY.-78f\">As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A <strong> titration curve <\/strong> is a graphical representation of the pH of a solution during a titration. The <strong> Figure <\/strong> below shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid and so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. This also corresponds to the color change of the indicator.<\/p>\r\n\r\n<div id=\"x-ck12-MGUyYjE0ZTlkMTcyYjIyZGFmYTcxN2NmYzUyMWRjYWM.-pdm\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NDk4NTgwOS02NC04LTQ.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213351\/20140811155917052567.png\" alt=\"Titration curves of strong acids and strong bases\" width=\"500\" height=\"292\" longdesc=\"A%20titration%20curve%20shows%20the%20pH%20changes%20that%20occur%20during%20the%20titration%20of%20an%20acid%20with%20a%20base.%20On%20the%20left%2C%20base%20is%20being%20added%20to%20acid.%20On%20the%20right%2C%20acid%20is%20being%20added%20to%20base.%20In%20both%20cases%2C%20the%20equivalence%20point%20is%20at%20pH%207.\" \/> Figure 2.\u00a0A titration curve shows the pH changes that occur during the titration of an acid with a base. On the left, base is being added to acid. On the right, acid is being added to base. In both cases, the equivalence point is at pH 7.[\/caption]\r\n\r\n<\/div>\r\n<p id=\"x-ck12-ZmYwYmZlY2MwM2I4YjRjOWZlZTdhOTM2Mzk0YzM3NGE.-ynn\">Titration curves can also be generated in the case of a weak acid-strong base titration or a strong base-weak acid titration. The general shape of the titration curve is the same, but the pH at the equivalence point is different. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point.<\/p>\r\n\r\n<div id=\"x-ck12-ZDRiNjBlNDIyMjAxMmI4YjJjMTE0NWViMDk2NzRmYmI.-c37\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NDk4NTgzOC0wMi0yMC01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213353\/20140811155917174940.png\" alt=\"Titration curve of a weak acid and strong base\" width=\"500\" height=\"408\" longdesc=\"Titration%20curve%20of%20weak%20acid%20and%20strong%20base.\" \/> Figure 3.\u00a0Titration curve of weak acid and strong base.[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ZjRlZmQ3ZjkzN2ZjMjdmZTA0ZTM4ZTU1ZWZhOGFmZTI.-b5o\">\r\n \t<li>Definitions are given for equivalence point, titration and indicator.<\/li>\r\n \t<li>The process for carrying out a titration is described.<\/li>\r\n \t<li>The process of calculating concentration from titration data is described and illustrated.<\/li>\r\n \t<li>Acid-base titration curves are described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Experiment<\/h4>\r\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-bys\">Watch the video at the link below and answer the following questions:<\/p>\r\nhttps:\/\/youtu.be\/g8jdCWC10vQ\r\n<ol id=\"x-ck12-YjQ4ODdkNzA2MGFjN2I0NmNjMzQ0OGU4ZWRhYTgzZDk.-p17\">\r\n \t<li>What is the indicator used?<\/li>\r\n \t<li>What color is it in acid solution?<\/li>\r\n \t<li>What is the glass tube called that contains the known concentration of sodium hydroxide?<\/li>\r\n \t<li>What other method can be used to determine the end-point of the titration?<\/li>\r\n<\/ol>\r\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Calculations<\/h4>\r\n<p id=\"x-ck12-ZjRmZTgxMGFkNmYzZWFhZGViYzllOGZjOWIxZDdhZjQ.-6g4\">Do the problems at the link below:<\/p>\r\n<p id=\"x-ck12-NTgwMjFjM2U4ZjM0ZTJlZGMwMDNlMDRmNTcwMDc5Mjk.-uss\"><a href=\"http:\/\/www.sophia.org\/acidbase-titration-calculations-concept\"> http:\/\/www.sophia.org\/acidbase-titration-calculations-concept<\/a><\/p>\r\n\r\n<h4>Titration Curves<\/h4>\r\n<p id=\"x-ck12-ZDA1NTE4MmE3ZWNhMTc1ZjZmZjlhNTc2YTQzZWI5ZTg.-zgs\">Read the material at the link below and answer the following questions:<\/p>\r\n<p id=\"x-ck12-N2JkZmVlNTZhNjA2MGM5MzRlMmRkYmM4MWJiOTNiODY.-n4z\"><a href=\"http:\/\/www.chemguide.co.uk\/physical\/acidbaseeqia\/phcurves.html\"> http:\/\/www.chemguide.co.uk\/physical\/acidbaseeqia\/phcurves.html <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-ODAwNWNkM2FlNDBjOGNiMzU0NTI4NDVlNmM5MmQ2ODE.-adr\">\r\n \t<li>Why is the equivalence point less than pH 7 for the titration of ammonia with HCl?<\/li>\r\n \t<li>Why is it difficult to do a titration of a weak acid and a weak base?<\/li>\r\n \t<li>Why do we get two inflection points for the titration of ethanedioic acid?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Experiment<\/h4>\r\n<ol id=\"x-ck12-MDllYjUyZGRiZjBhMjVlMjExZTNmMTZjZGIxNjk5MWI.-th6\">\r\n \t<li>What is the standard solution?<\/li>\r\n \t<li>How do you know you have reached the end-point?<\/li>\r\n \t<li>What is the reaction that occurs during a titration?<\/li>\r\n<\/ol>\r\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Calculations<\/h4>\r\n<ol>\r\n \t<li>What assumption is made about the amounts of materials at the neutral point?<\/li>\r\n \t<li>What is different about the calculation using sulfuric acid?<\/li>\r\n \t<li>Why is the mole ratio important?<\/li>\r\n<\/ol>\r\n<h4>Titration Curves<\/h4>\r\n<ol>\r\n \t<li>What does a titration curve tell us?<\/li>\r\n \t<li>At what pH are the moles of acid and base equal?<\/li>\r\n \t<li>Is the equivalence point for a weak acid-strong base titration the same as for a strong-acid-strong base titration?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-YjMxMDhjZmM2MTQwNDBlZGUyMWE2NmJmNTA2Y2M4YzQ.-ydw\">\r\n \t<li><strong> end point: <\/strong> The point at which the indicator changes color.<\/li>\r\n \t<li><strong> equivalence point: <\/strong> The point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.<\/li>\r\n \t<li><strong> indicator: <\/strong> A substance that has a distinctly different color when in an acidic or basic solution.<\/li>\r\n \t<li><strong> standard solution: <\/strong> The solution in a titration whose concentration is known.<\/li>\r\n \t<li><strong> titration: <\/strong> An experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration.<\/li>\r\n \t<li><strong style=\"line-height: 1.5;\">titration curve: <\/strong><span style=\"line-height: 1.5;\"> A graphical representation of the pH of a solution during a titration.<\/span><\/li>\r\n<\/ul>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>User:GeorgHH\/Wikimedia Commons.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Zapfs%C3%A4ule_044_3.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Zapfs%C3%A4ule_044_3.jpg <\/a>.<\/li>\r\n \t<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg <\/a>.<\/li>\r\n \t<li>User:Phanton\/Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Decorative_Soaps.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Decorative_Soaps.jpg <\/a>.<\/li>\r\n \t<li>William Holl the Younger (1807-1871) after Frans Hals.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:William_Holl_the_Younger06.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:William_Holl_the_Younger06.jpg <\/a>.<\/li>\r\n \t<li>Laura Guerin. CK-12 Foundation.<\/li>\r\n \t<li>Laura Guerin. CK-12 Foundation.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define equivalence point.<\/li>\n<li>Describe how to perform a titration experiment.<\/li>\n<li>Perform calculations to determine concentration of unknown acid or base.<\/li>\n<li>Describe titration curves of acid-base neutralization reactions.<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"x-ck12-VGl0cmF0aW9uIEV4cGVyaW1lbnQ.\">Titration Experiment<\/h2>\n<div class=\"textbox examples\">\n<h3><strong>Didn\u2019t that used to be French fries? <\/strong><\/h3>\n<p id=\"x-ck12-YjhiYmY3YzQ0ZDUxMWY3NjA4NzI0MzY4YTM4NjQxYjU.-tik\">A lot of research is going on these days involving the development of biodiesel fuels. Often this material can be made from used vegetable oils. The vegetable oil is treated with lye to create the biofuel. In the oils is a variable amount of acid that needs to be determined so the workers will know how much lye to add to make the final fuel. Before the lye is added, the native vegetable oil is titrated to find out how much free acid is present. Then the amount of lye added can be adjusted to take into account the amount needed to neutralize these free acids.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213332\/20140811155916481045.jpeg\" alt=\"Biodiesel synthesis requires the amount of acid to be determined before adding lye\" width=\"250\" \/><\/p>\n<\/div>\n<p id=\"x-ck12-YjNmZmVjMDMyNDgyZDA0ZDQxNjRjYWNkOGU1ODlkZDA.-owf\">In the neutralization of hydrochloric acid by sodium hydroxide, the mole ratio of acid to base is 1:1.<\/p>\n<p id=\"x-ck12-qw6\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211839\/4b5dd661a2a365a9ee36f5fc7737474b.png\" alt=\"text{HCl}(aq)+text{NaOH}(aq) rightarrow text{NaCl}(aq)+text{H}_2text{O}(l)\" width=\"343\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-M2Q1Zjk0MDdmY2FkYjM1Mzg2NGM0NDA0MTVkZjgyZWI.-tec\">One mole of HCl would be fully neutralized by one mole of NaOH. If instead the hydrochloric acid was reacted with barium hydroxide, the mole ratio would be 2:1.<\/p>\n<p id=\"x-ck12-0bt\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213334\/63259ec0f75dffbad96f90f8350cf63e.png\" alt=\"2text{HCl}(aq)+text{Ba}(text{OH})_2(aq) rightarrow text{BaCl}_2(aq)+2text{H}_2text{O}(l)\" width=\"387\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-MDRhNGI4MjljZDA5NzRiMDZkODEzZTBhYTUwMTVkMDA.-sda\">Now two moles of HCl would be required to neutralize one mole of Ba(OH) <sub> 2 <\/sub> . The mole ratio insures that the number of moles of H <sup> + <\/sup> ions supplied by the acid is equal to the number of OH <sup> \u2212 <\/sup> ions supplied by the base. This must be the case for neutralization to occur. The <strong> equivalence point <\/strong> is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.<\/p>\n<p id=\"x-ck12-ZDM5ZWM0MGFjN2YzOWI4N2I2NmQ1MmY1YTFkZTBjN2Q.-u6p\">In the laboratory, it is useful to have an experiment where the unknown concentration of an acid or a base can be determined. This can be accomplished by performing a controlled neutralization reaction. A <strong> titration <\/strong> is an experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. Many titrations are acid-base neutralization reactions, though other types of titrations can also be performed.<\/p>\n<p id=\"x-ck12-NmIxNzAxMDYxNzRjM2EzZDdiNjZmMDY4NTI1MjJkOWM.-xan\">In order to perform an acid-base titration, the chemist must have a way to visually detect that the neutralization reaction has occurred. An <strong> indicator <\/strong> is a substance that has a distinctly different color when in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein. Solutions in which a few drops of phenolphthalein have been added turn from colorless to brilliant pink as the solution turns from acidic to basic. The steps in a titration reaction are outlined below.<\/p>\n<ol id=\"x-ck12-YzdiMThlZjIyZTlkNTY2MjU1N2Y1ZDkzY2E4ZjYxNTI.-npb\">\n<li>A measured volume of an acid of unknown concentration is added to an Erlenmeyer flask.<\/li>\n<li>Several drops of an indicator are added to the acid and mixed by swirling the flask.<\/li>\n<li>A buret is filled with the base solution of known molarity.<\/li>\n<li>The stopcock of the buret is opened and base is slowly added to the acid while the flask is constantly swirled to insure mixing. The stopcock is closed at the exact point at which the indicator just changes color.<\/li>\n<\/ol>\n<div id=\"x-ck12-YWE5OThkMjA3NjllNDBiNjBhZDAzNzYyZWM3ZDc5Y2M.-xpv\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<div style=\"width: 202px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NDk4MjMwMi00MS0yMS01Ng..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213335\/20140811155916680904.jpeg\" alt=\"Phenolphthalein turns pink in basic solutions\" width=\"192\" height=\"172\" longdesc=\"Phenolphthalein%20in%20basic%20solution.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1.\u00a0Phenolphthalein in basic solution.<\/p>\n<\/div>\n<p>The <strong> standard solution <\/strong> is the solution in a titration whose concentration is known. In the titration described above the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly so that the point at which the indicator changes color can be found accurately.<\/p>\n<p>The <strong> end point <\/strong> of a titration is the point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by a faint pink color.<\/p>\n<\/div>\n<h2 id=\"x-ck12-VGl0cmF0aW9uIENhbGN1bGF0aW9ucw..\">Titration Calculations<\/h2>\n<div class=\"textbox examples\">\n<h3>How is soap made?<\/h3>\n<p id=\"x-ck12-NGIwNDYzMmJjOGMyY2IxYWYxZGVhOWQ3NTY1YWUxNWQ.-hm3\"><span class=\"x-ck12-img-inline\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213337\/20140811155916755515.jpeg\" alt=\"The calculation of saponification number in the production of soap is typically found through titration\" width=\"233\" height=\"175\" \/><\/span><\/p>\n<p id=\"x-ck12-ZWViZmZlM2Y5MTRiMjE2Nzc3NTg1YjlmMjQ3YWZmMmQ.-ris\">The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product.<\/p>\n<p>The fat is heated with a known amount of base (usually NaOH or KOH). After hydrolysis is complete, the left-over base is titrated to determine how much was needed to hydrolyze the fat sample.<\/p>\n<\/div>\n<h3>Titration Calculations<\/h3>\n<p id=\"x-ck12-NGY1MTMyOGZmZmM1ZjI4ZjE2NTMwNThiOTJhMWM1MjQ.-alb\">At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.<\/p>\n<p>moles acid = moles base<\/p>\n<p id=\"x-ck12-YzMwMTE1ZjBiNDY2OWFjNjhiMTQ3MjFjZjc0YWUxZDg.-zkw\">Recall that the molarity (<em>M<\/em>)\u00a0of a solution is defined as the moles of the solute divided by the liters of solution (<em>L<\/em>). So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.<\/p>\n<p id=\"x-ck12-jc8\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213341\/7617be94d4cc9da7450f0ec91ebdf9f4.png\" alt=\"text{moles solute} = M times L\" width=\"173\" height=\"14\" \/><\/p>\n<p id=\"x-ck12-ZjU1Yjk3NGY3ZmI5MWE1NmVlMDQwZTJhM2ZhNzRjZGE.-dlx\">We can then set the moles of acid equal to the moles of base.<\/p>\n<p id=\"x-ck12-qe5\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213342\/724089f5770bbe6377231cc383015a42.png\" alt=\"M_A times V_A=M_B times V_B\" width=\"165\" height=\"15\" \/><\/p>\n<p id=\"x-ck12-ODlhZDkxNTVmNTk5OTU3NjlhOWU3ZDc2ODhlNGE2NWM.-eqm\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213342\/f094c5600c0d02a8ec44fbd0b26935d0.png\" alt=\"M_A\" width=\"27\" height=\"15\" \/> \u00a0is the molarity of the acid, while\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/43f078a33f698a7025f295f338e19214.png\" alt=\"M_B\" width=\"27\" height=\"15\" \/> is the molarity of the base.\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/7badc172c5b50023ee2e75b381450c4d.png\" alt=\"V_A\" width=\"20\" height=\"15\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213343\/f2d1aa5f5039db0bdf46158eb1cde675.png\" alt=\"V_B\" width=\"20\" height=\"15\" \/> are the volumes of the acid and base, respectively.<\/p>\n<p id=\"x-ck12-MDc4NTI2N2NjZThmNTJlZDg2NGVhMDI5ZWM1MjVlYzg.-akd\">Suppose that a titration is performed and 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated against 15.00 mL of HCl of unknown concentration. The above equation can be used to solve for the molarity of the acid.<\/p>\n<p id=\"x-ck12-j3q\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213344\/815ee7cbc6d01dad16fa9e7e3ee6821f.png\" alt=\"M_A=frac{M_B times V_B}{V_A}=frac{0.500 text{M} times 20.70 text{mL}}{15.00 text{mL}}=0.690 text{M}\" width=\"400\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-NDljOTk4Y2U1M2M1MTI4ZGQwOGM4YmE2NzQ0OWNmMjc.-jug\">The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.<\/p>\n<p id=\"x-ck12-MDdiZTNmNWNmMTEyNTBlMDkzMDVhNjNkY2ZhY2JhM2Y.-voc\">The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The sample problem below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.<\/p>\n<div class=\"textbox shaded\">\n<h4>Sample Problem: Titration<\/h4>\n<p id=\"x-ck12-OGY1YmRkMmU3Mjg1MTdjNWYyOWEzN2M5OTFmMDNlYjE.-qbm\">In a titration of sulfuric acid against sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H<sub>2<\/sub>SO<sub>4<\/sub>. Calculate the molarity of the sulfuric acid.<\/p>\n<p id=\"x-ck12-OGM1ZTljNzdiMjc4MjQyMjY3OTljNDI4OTk2ZmY0MTE.-o8g\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-byj\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-OTA4MjA5ZmZlYzljZGNlZGZlNTZlZGI4MDNmM2JmNWM.-vuc\">\n<li>molarity NaOH = 0.250 M<\/li>\n<li>volume NaOH = 32.20 mL<\/li>\n<li>volume H <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 26.60 mL<\/li>\n<\/ul>\n<p id=\"x-ck12-NzNiNGU0YWI4NDkzN2MxNTQyYTUzOTViMzRiMzY4ZTI.-rwf\"><span class=\"x-ck12-underline\"> Unkonwn <\/span><\/p>\n<ul id=\"x-ck12-YmExOGMzZGE4Y2JhNzVjYmVhM2Q4ZGRiNTg0YzZlNmU.-kk3\">\n<li>molarity H <sub> 2 <\/sub> SO <sub> 4 <\/sub> = ?<\/li>\n<\/ul>\n<p id=\"x-ck12-ZGRmMTAxMTkwNTZhMjBjZjI2MTE1Y2QwZDM5ODEwZjE.-kct\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213345\/377340a89ea67f537ffdfbbde8ed62ec.png\" alt=\"text{equation} qquad text{H}_2 text{SO}_4(aq)+2text{NaOH}(aq) rightarrow text{Na}_2text{SO}_4(aq)+2text{H}_2text{O}(l)\" width=\"504\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-YzM2MDJiMTY0ZDY2NzlmMDc2OTQzODlmOTdmZTlmOTk.-yli\">First determine the moles of NaOH in the reaction. From the mole ratio, calculate the moles of H <sub> 2 <\/sub> SO <sub> 4 <\/sub> that reacted. Finally, divide the moles H <sub> 2 <\/sub> SO <sub> 4 <\/sub> by its volume to get the molarity.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-mu2\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-qea\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213347\/75ecadf37aa4f4b55d37d61178ad6acb.png\" alt=\"&amp; text{mol NaOH}=M times L=0.250 text{M} times 0.03220 text{L}=8.05 times 10^{-3} text{mol} NaOH \\&amp; 8.05 times 10^{-3} text{mol NaOH} times frac{1 text{mol H}_2text{SO}_4}{2 text{mol NaOH}}=4.03 times 10^{-3} text{mol H}_2text{SO}_4 \\&amp; frac{4.03 times 10^{-3} text{mol H}_2text{SO}_4}{0.02660 text{L}}=0.151 text{M H}_2text{SO}_4\" width=\"558\" height=\"113\" \/><\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-mbl\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NzZkMDhiNzI4MTAzMmIxODc0NWNlNmVkY2Y2ODc5NDM.-vfg\">The volume of H<sub>2<\/sub>SO<sub>4 <\/sub> required is smaller than the volume of NaOH because of the two hydrogen ions contributed by each molecule.<\/p>\n<\/div>\n<h2 id=\"x-ck12-VGl0cmF0aW9uIEN1cnZlcw..\">Titration Curves<\/h2>\n<p id=\"x-ck12-MDMzNjgzOGVhYjQ0Yjc0MTkxMWZiYjQ1YmM3YmY1OGI.-ykm\">\n<div class=\"textbox examples\">\n<h3><strong>Where did graphs come from? <\/strong><\/h3>\n<p id=\"x-ck12-OTIyOWY2NDg5MmM2NjM2YmIzZTI1ZWVjNDhmYmRlMzk.-bcm\">The\u00a0 <em>x\u2013y<\/em>\u00a0plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596\u20131650). His studies in mathematics led him to develop what was known as \u201cCartesian geometry,\u201d including the idea of our current graphs. The coordinates are often referred to as Cartesian coordinates.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213349\/20140811155916885001.jpeg\" alt=\"Cartesian graphs are often used to represent the course of a titration\" width=\"200\" \/><\/p>\n<\/div>\n<h3>Titration Curves<\/h3>\n<p id=\"x-ck12-ZmIzYjk4ODcwZDI3MzQ5N2JhZmIwMzY0MDIyYTU3OTY.-78f\">As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A <strong> titration curve <\/strong> is a graphical representation of the pH of a solution during a titration. The <strong> Figure <\/strong> below shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid and so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. This also corresponds to the color change of the indicator.<\/p>\n<div id=\"x-ck12-MGUyYjE0ZTlkMTcyYjIyZGFmYTcxN2NmYzUyMWRjYWM.-pdm\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NDk4NTgwOS02NC04LTQ.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213351\/20140811155917052567.png\" alt=\"Titration curves of strong acids and strong bases\" width=\"500\" height=\"292\" longdesc=\"A%20titration%20curve%20shows%20the%20pH%20changes%20that%20occur%20during%20the%20titration%20of%20an%20acid%20with%20a%20base.%20On%20the%20left%2C%20base%20is%20being%20added%20to%20acid.%20On%20the%20right%2C%20acid%20is%20being%20added%20to%20base.%20In%20both%20cases%2C%20the%20equivalence%20point%20is%20at%20pH%207.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2.\u00a0A titration curve shows the pH changes that occur during the titration of an acid with a base. On the left, base is being added to acid. On the right, acid is being added to base. In both cases, the equivalence point is at pH 7.<\/p>\n<\/div>\n<\/div>\n<p id=\"x-ck12-ZmYwYmZlY2MwM2I4YjRjOWZlZTdhOTM2Mzk0YzM3NGE.-ynn\">Titration curves can also be generated in the case of a weak acid-strong base titration or a strong base-weak acid titration. The general shape of the titration curve is the same, but the pH at the equivalence point is different. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point.<\/p>\n<div id=\"x-ck12-ZDRiNjBlNDIyMjAxMmI4YjJjMTE0NWViMDk2NzRmYmI.-c37\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NDk4NTgzOC0wMi0yMC01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213353\/20140811155917174940.png\" alt=\"Titration curve of a weak acid and strong base\" width=\"500\" height=\"408\" longdesc=\"Titration%20curve%20of%20weak%20acid%20and%20strong%20base.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3.\u00a0Titration curve of weak acid and strong base.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ZjRlZmQ3ZjkzN2ZjMjdmZTA0ZTM4ZTU1ZWZhOGFmZTI.-b5o\">\n<li>Definitions are given for equivalence point, titration and indicator.<\/li>\n<li>The process for carrying out a titration is described.<\/li>\n<li>The process of calculating concentration from titration data is described and illustrated.<\/li>\n<li>Acid-base titration curves are described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Experiment<\/h4>\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-bys\">Watch the video at the link below and answer the following questions:<\/p>\n<p>https:\/\/youtu.be\/g8jdCWC10vQ<\/p>\n<ol id=\"x-ck12-YjQ4ODdkNzA2MGFjN2I0NmNjMzQ0OGU4ZWRhYTgzZDk.-p17\">\n<li>What is the indicator used?<\/li>\n<li>What color is it in acid solution?<\/li>\n<li>What is the glass tube called that contains the known concentration of sodium hydroxide?<\/li>\n<li>What other method can be used to determine the end-point of the titration?<\/li>\n<\/ol>\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Calculations<\/h4>\n<p id=\"x-ck12-ZjRmZTgxMGFkNmYzZWFhZGViYzllOGZjOWIxZDdhZjQ.-6g4\">Do the problems at the link below:<\/p>\n<p id=\"x-ck12-NTgwMjFjM2U4ZjM0ZTJlZGMwMDNlMDRmNTcwMDc5Mjk.-uss\"><a href=\"http:\/\/www.sophia.org\/acidbase-titration-calculations-concept\"> http:\/\/www.sophia.org\/acidbase-titration-calculations-concept<\/a><\/p>\n<h4>Titration Curves<\/h4>\n<p id=\"x-ck12-ZDA1NTE4MmE3ZWNhMTc1ZjZmZjlhNTc2YTQzZWI5ZTg.-zgs\">Read the material at the link below and answer the following questions:<\/p>\n<p id=\"x-ck12-N2JkZmVlNTZhNjA2MGM5MzRlMmRkYmM4MWJiOTNiODY.-n4z\"><a href=\"http:\/\/www.chemguide.co.uk\/physical\/acidbaseeqia\/phcurves.html\"> http:\/\/www.chemguide.co.uk\/physical\/acidbaseeqia\/phcurves.html <\/a><\/p>\n<ol id=\"x-ck12-ODAwNWNkM2FlNDBjOGNiMzU0NTI4NDVlNmM5MmQ2ODE.-adr\">\n<li>Why is the equivalence point less than pH 7 for the titration of ammonia with HCl?<\/li>\n<li>Why is it difficult to do a titration of a weak acid and a weak base?<\/li>\n<li>Why do we get two inflection points for the titration of ethanedioic acid?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Experiment<\/h4>\n<ol id=\"x-ck12-MDllYjUyZGRiZjBhMjVlMjExZTNmMTZjZGIxNjk5MWI.-th6\">\n<li>What is the standard solution?<\/li>\n<li>How do you know you have reached the end-point?<\/li>\n<li>What is the reaction that occurs during a titration?<\/li>\n<\/ol>\n<h4 id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-9am\">Titration Calculations<\/h4>\n<ol>\n<li>What assumption is made about the amounts of materials at the neutral point?<\/li>\n<li>What is different about the calculation using sulfuric acid?<\/li>\n<li>Why is the mole ratio important?<\/li>\n<\/ol>\n<h4>Titration Curves<\/h4>\n<ol>\n<li>What does a titration curve tell us?<\/li>\n<li>At what pH are the moles of acid and base equal?<\/li>\n<li>Is the equivalence point for a weak acid-strong base titration the same as for a strong-acid-strong base titration?<\/li>\n<\/ol>\n<\/div>\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-YjMxMDhjZmM2MTQwNDBlZGUyMWE2NmJmNTA2Y2M4YzQ.-ydw\">\n<li><strong> end point: <\/strong> The point at which the indicator changes color.<\/li>\n<li><strong> equivalence point: <\/strong> The point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.<\/li>\n<li><strong> indicator: <\/strong> A substance that has a distinctly different color when in an acidic or basic solution.<\/li>\n<li><strong> standard solution: <\/strong> The solution in a titration whose concentration is known.<\/li>\n<li><strong> titration: <\/strong> An experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration.<\/li>\n<li><strong style=\"line-height: 1.5;\">titration curve: <\/strong><span style=\"line-height: 1.5;\"> A graphical representation of the pH of a solution during a titration.<\/span><\/li>\n<\/ul>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>User:GeorgHH\/Wikimedia Commons.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Zapfs%C3%A4ule_044_3.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Zapfs%C3%A4ule_044_3.jpg <\/a>.<\/li>\n<li>Ben Mills (Wikimedia: Benjah-bmm27). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Phenolphthalein-at-pH-9.jpg <\/a>.<\/li>\n<li>User:Phanton\/Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Decorative_Soaps.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Decorative_Soaps.jpg <\/a>.<\/li>\n<li>William Holl the Younger (1807-1871) after Frans Hals.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:William_Holl_the_Younger06.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:William_Holl_the_Younger06.jpg <\/a>.<\/li>\n<li>Laura Guerin. CK-12 Foundation.<\/li>\n<li>Laura Guerin. CK-12 Foundation.<\/li>\n<\/ol>\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-2823\">\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>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/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><\/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":17,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2823","chapter","type-chapter","status-publish","hentry"],"part":2342,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2823","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2823\/revisions"}],"predecessor-version":[{"id":3046,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2823\/revisions\/3046"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2342"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2823\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2823"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2823"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2823"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2823"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}