{"id":7622,"date":"2021-06-08T21:57:04","date_gmt":"2021-06-08T21:57:04","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/acid-base-titrations\/"},"modified":"2023-09-25T21:05:25","modified_gmt":"2023-09-25T21:05:25","slug":"acid-base-titrations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/acid-base-titrations\/","title":{"raw":"Acid-Base Titrations","rendered":"Acid-Base Titrations"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Describe a titration experiment.<\/li>\r\n \t
  2. Explain what an indicator does.<\/li>\r\n \t
  3. Perform a titration calculation correctly.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nThe 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 titration<\/strong>. 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.\r\n\r\nIn 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 titrant<\/strong>) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the analyte<\/strong>) 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 Figure 12.1 \"Equipment for Titrations\"). 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 equivalence point<\/strong>; 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.\r\n\r\n[caption id=\"attachment_7616\" align=\"aligncenter\" width=\"200\"]\"\" Figure 12.1 \"Equipment for Titrations.\" A burette is a type of liquid dispensing system that can accurately indicate the volume of liquid dispensed.[\/caption]\r\n\r\nFor 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:\r\n

    [latex]\\#\\text{ mol HCl}=(0.02566\\text{ L})(0.1078\\text{ M})=0.002766\\text{ mol HCl}[\/latex]<\/p>\r\nWe also have the balanced chemical reaction between HCl and NaOH:\r\n

    HCl +\u00a0NaOH \u2192\u00a0NaCl +\u00a0H2<\/sub>O<\/p>\r\nSo we can construct a conversion factor to convert to number of moles of NaOH reacted:\r\n

    [latex]0.002766\\text{ \\cancel{mol HCl}}\\times\\dfrac{1\\text{ mol NaOH}}{1\\text{ \\cancel{mol HCl}}}=0.002766\\text{ mol NaOH}[\/latex]<\/p>\r\nThen we convert this amount to mass, using the molar mass of NaOH (40.00 g\/mol):\r\n

    [latex]0.002766\\text{ \\cancel{mol NaOH}}\\times\\dfrac{40.00\\text{ g NaOH}}{1\\text{ \\cancel{mol NaOH}}}=0.1106\\text{ g NaOH}[\/latex]<\/p>\r\nThis is type of calculation is performed as part of a titration.\r\n

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    Example 12.1<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat mass of Ca(OH)2<\/sub> is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO3<\/sub>? The balanced chemical equation is as follows:\r\n

    2HNO3<\/sub> + Ca(OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + 2H2<\/sub>O<\/p>\r\nSolution<\/em>\r\nIn litres, the volume is 0.04402 L. We calculate the number of moles of titrant:\r\n

    [latex]\\#\\text{ moles }\\ce{HNO3}=(0.04402\\text{ L})(0.0885\\text{ M})=0.00390\\text{ mol }\\ce{HNO3}[\/latex]<\/p>\r\nUsing the balanced chemical equation, we can determine the number of moles of Ca(OH)2<\/sub> present in the analyte:\r\n

    [latex]0.00390\\cancel{\\text{ mol }\\ce{HNO3}}\\times\\dfrac{1\\text{ mol }\\ce{Ca(OH)2}}{2\\cancel{\\text{ mol }\\ce{HNO3}}}=0.00195\\text{ mol }\\ce{Ca(OH)2}[\/latex]<\/p>\r\nThen we convert this to a mass using the molar mass of Ca(OH)2<\/sub>:\r\n

    [latex]0.00195\\cancel{\\text{ mol }\\ce{Ca(OH)2}}\\times \\dfrac{74.1\\text{ g }\\ce{Ca(OH)2}}{1\\cancel{\\text{ mol }\\ce{Ca(OH)2}}}=0.144\\text{ g }\\ce{Ca(OH)2}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\n\r\nWhat mass of H2<\/sub>C2<\/sub>O4<\/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:\r\n

    H2<\/sub>C2<\/sub>O4<\/sub> + 2NaOH \u2192 Na2<\/sub>C2<\/sub>O4<\/sub> + 2H2<\/sub>O<\/p>\r\nAnswer<\/em>\r\n0.182 g\r\n\r\n<\/div>\r\n<\/div>\r\nHow does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an indicator<\/strong>, a substance that changes colour depending on the acidity or basicity of the solution. Because different indicators change colours at different levels of acidity, choosing the correct one is important in performing an accurate titration.\r\n\r\nFor more information, watch this video: \"Titration\"<\/a> by Jessie Key.\r\n

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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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