{"id":7608,"date":"2021-06-08T21:57:00","date_gmt":"2021-06-08T21:57:00","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/quantitative-units-of-concentration\/"},"modified":"2021-10-04T22:33:15","modified_gmt":"2021-10-04T22:33:15","slug":"quantitative-units-of-concentration","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/quantitative-units-of-concentration\/","title":{"raw":"Quantitative Units of Concentration","rendered":"Quantitative Units of Concentration"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Learn to determine specific concentrations with several common units.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nRather than qualitative terms (see the later section \"Some Definitions\"<\/a>), we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration.\r\n\r\nMolarity<\/strong> (M) is defined as the number of moles of solute divided by the number of litres of solution:\r\n

    [latex]\\text{molarity}=\\dfrac{\\text{moles of solute}}{\\text{litres of solution}}[\/latex]<\/p>\r\nwhich can be simplified as\r\n

    [latex]\\text{M}=\\dfrac{\\text{mol}}{\\text{L}},\\text{ or mol\/L}[\/latex]<\/p>\r\nAs with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity.\r\n\r\nFor example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows:\r\n

    [latex]\\text{molarity}=\\dfrac{0.24\\text{ mol NaOH}}{0.500\\text{ L}}=0.48\\text{ M NaOH}[\/latex]<\/p>\r\nThe concentration of the solution is 0.48 M, which is spoken as \u201czero point forty-eight molarity\u201d or \u201czero point forty-eight molar.\u201d If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? First, convert the mass of solute to moles using the molar mass of HCl (36.5 g\/mol):\r\n

    [latex]22.4\\text{ \\cancel{g HCl}}\\times \\dfrac{1\\text{ mol HCl}}{36.5\\text{ \\cancel{g HCl}}}=0.614\\text{ mol HCl}[\/latex]<\/p>\r\nNow we can use the definition of molarity to determine a concentration:\r\n

    [latex]\\text{M}=\\dfrac{0.614\\text{ mol HCl}}{1.56\\text{ L}}=0.394\\text{ M}[\/latex]<\/p>\r\n\r\n

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    Example 11.11<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the molarity of a solution made when 32.7 g of NaOH are dissolved to make 445 mL of solution?\r\n\r\nSolution<\/em>\r\nTo use the definition of molarity, both quantities must be converted to the proper units. First, convert the volume units from millilitres to litres:\r\n

    [latex]445\\text{ \\cancel{mL}}\\times \\dfrac{1\\text{ L}}{1000\\text{ \\cancel{mL}}}=0.445\\text{ L}[\/latex]<\/p>\r\nNow we convert the amount of solute to moles, using the molar mass of NaOH, which is 40.0 g\/mol:\r\n

    [latex]32.7\\text{ \\cancel{g NaOH}}\\times \\dfrac{1\\text{ mol NaOH}}{40.0\\text{ \\cancel{g NaOH}}}=0.818\\text{ mol NaOH}[\/latex]<\/p>\r\nNow we can use the definition of molarity to determine the molar concentration:\r\n

    [latex]\\text{M}=\\dfrac{0.818\\text{ mol NaOH}}{0.445\\text{ L}}=1.84\\text{ M NaOH}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat is the molarity of a solution made when 66.2 g of C6<\/sub>H12<\/sub>O6<\/sub> are dissolved to make 235 mL of solution?\r\n\r\nAnswer<\/em>\r\n1.57 M\r\n\r\n<\/div>\r\n<\/div>\r\nThe definition of molarity can be used to determine the amount of solute or the volume of solution, if the other information is given. Example 11.12 illustrates this situation.\r\n

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    Example 11.12<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nHow many moles of solute are present in 0.108 L of a 0.887 M NaCl solution?\r\n\r\nSolution<\/em>\r\n\r\nWe know the volume and the molarity; we can use the definition of molarity to mathematically solve for the amount in moles. Substituting the quantities into the definition of molarity:\r\n

    [latex]0.887\\text{ M}=\\dfrac{1\\text{ mol NaCl}}{0.108\\text{ L}}[\/latex]<\/p>\r\nWe multiply the 0.108 L over to the other side of the equation and multiply the units together; \u201cmolarity \u00d7 litres\u201d equals moles, according to the definition of molarity. So:\r\n

    [latex]1\\text{ mol NaCl}=(0.887\\text{ M})(0.108\\text{ L})=0.0958\\text{ mol}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nHow many moles of solute are present in 225 mL of a 1.44 M CaCl2<\/sub> solution?\r\n\r\nAnswer<\/em>\r\n0.324 mol\r\n\r\n<\/div>\r\n<\/div>\r\nIf you need to determine volume, remember the rule that the unknown quantity must be by itself and in the numerator to determine the correct answer. Thus rearrangement of the definition of molarity is required.\r\n

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    Example 11.13<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat volume of a 2.33 M NaNO3<\/sub> solution is needed to obtain 0.222 mol of solute?\r\n\r\nSolution<\/em>\r\n\r\nUsing the definition of molarity, we have:\r\n

    [latex]2.33\\text{ M}=\\dfrac{0.222\\text{ mol}}{\\text{L}}[\/latex]<\/p>\r\nTo solve for the number of litres, we bring the 2.33 M over to the right into the denominator, and the number of litres over to the left in the numerator. We now have:\r\n

    [latex]\\text{L}=\\dfrac{0.222\\text{ mol}}{2.33\\text{ M}}[\/latex]<\/p>\r\nDividing, the volume is 0.0953 L = 95.3 mL.\r\n\r\nTest Yourself<\/em>\r\n\r\nWhat volume of a 0.570 M K2<\/sub>SO4<\/sub> solution is needed to obtain 0.872 mol of solute?\r\n\r\nAnswer<\/em>\r\n1.53 L\r\n\r\n<\/div>\r\n<\/div>\r\nA similar unit of concentration is molality<\/strong>\u00a0(m<\/em>), which is defined as the number of moles of solute per kilogram of solvent, not per litre of solution:\r\n

    [latex]\\text{molality}=\\dfrac{\\text{moles solute}}{\\text{kilograms solvent}}[\/latex]<\/p>\r\nMathematical manipulation of molality is the same as with molarity.\r\n\r\nAnother way to specify an amount is percentage composition by mass<\/strong>\u00a0(or mass percentage<\/em>, % m\/m). It is defined as follows:\r\n

    [latex]\\%\\text{ m\/m}=\\dfrac{\\text{mass of solute}}{\\text{mass of entire sample}}\\times 100\\%[\/latex]<\/p>\r\nIt is not uncommon to see this unit used on commercial products.\r\n

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    Example 11.14<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the mass percentage of Fe in a piece of metal with 87.9 g of Fe in a 113 g sample?\r\n\r\nSolution<\/em>\r\nUsing the definition of mass percentage, we have:\r\n

    [latex]\\%\\text{ m\/m}=\\dfrac{87.9\\text{ g Fe}}{113\\text{ g sample}}\\times 100\\%=77.8\\%\\text{ Fe}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat is the mass percentage of H2<\/sub>O2<\/sub> in a solution with 1.67 g of H2<\/sub>O2<\/sub> in a 55.5 g sample?\r\n\r\nAnswer<\/em>\r\n3.01%\r\n\r\n<\/div>\r\n<\/div>\r\nRelated concentration units are parts per thousand (ppth)<\/strong>, parts per million (ppm)<\/strong>, and parts per billion (ppb)<\/strong>. Parts per thousand is defined as follows:\r\n

    [latex]\\text{ppth}=\\dfrac{\\text{mass of solute}}{\\text{mass of sample}}\\times 1,000[\/latex]<\/p>\r\nThere are similar definitions for parts per million and parts per billion:\r\n

    [latex]\\begin{array}{rcl}\r\n\\text{ppm}&=&\\dfrac{\\text{mass of solute}}{\\text{mass of sample}}\\times 1,000,000 \\\\ \\\\\r\n\\text{ppb}&=&\\dfrac{\\text{mass of solute}}{\\text{mass of sample}}\\times 1,000,000,000 \\\\\r\n\\end{array}[\/latex]<\/p>\r\nEach unit is used for progressively lower and lower concentrations. The two masses must be expressed in the same unit of mass, so conversions may be necessary.\r\n

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    Example 11.15<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nIf there is 0.6 g of Pb present in 277 g of solution, what is the Pb concentration in parts per thousand?\r\n\r\nSolution<\/em>\r\nUse the definition of parts per thousand to determine the concentration. Substituting:\r\n

    [latex]\\dfrac{0.6\\text{ g Pb}}{277\\text{ g solution}}\\times 1000=2.17\\text{ ppth}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nIf there is 0.551 mg of As in 348 g of solution, what is the As concentration in ppm?\r\n\r\nAnswer<\/em>\r\n1.58 ppm\r\n\r\n<\/div>\r\n<\/div>\r\nAs with molarity and molality, algebraic rearrangements may be necessary to answer certain questions.\r\n

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    Example 11.16<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nThe concentration of Cl\u2013<\/sup> ion in a sample of H2<\/sub>O is 15.0 ppm. What mass of Cl\u2013<\/sup> ion is present in 240.0 mL of H2<\/sub>O, which has a density of 1.00 g\/mL?\r\n\r\nSolution<\/em>\r\nFirst, use the density of H2<\/sub>O to determine the mass of the sample:\r\n

    [latex]240.0\\text{ \\cancel{mL}}\\times\\dfrac{1.00\\text{ g}}{\\text{ \\cancel{mL}}}=240.0\\text{ g}[\/latex]<\/p>\r\nNow we can use the definition of ppm:\r\n

    [latex]15.0\\text{ ppm}=\\dfrac{\\text{mass solute}}{240.0\\text{ g solution}}\\times 1,000,000[\/latex]<\/p>\r\nRearranging to solve for the mass of solute:\r\n

    [latex]\\text{mass solute}=\\dfrac{(15.0\\text{ ppm})(240.0\\text{ g solution})}{1,000,000}=0.0036\\text{ g}=3.6\\text{ mg}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nThe concentration of Fe3+<\/sup> ion in a sample of H2<\/sub>O is 335.0 ppm. What mass of Fe3+<\/sup> ion is present in 3,450 mL of H2<\/sub>O, which has a density of 1.00 g\/mL?\r\n\r\nAnswer<\/em>\r\n1.16 g\r\n\r\n<\/div>\r\n<\/div>\r\nFor ionic solutions, we need to differentiate between the concentration of the salt versus the concentration of each individual ion. Because the ions in ionic compounds go their own way when a compound is dissolved in a solution, the resulting concentration of the ion may be different from the concentration of the complete salt. For example, if 1 M NaCl were prepared, the solution could also be described as a solution of 1 M Na+<\/sup>(aq) and 1 M Cl\u2212<\/sup>(aq) because there is one Na+<\/sup> ion and one Cl\u2212<\/sup> ion per formula unit of the salt. However, if the solution were 1 M CaCl2<\/sub>, there are two Cl\u2212<\/sup>(aq) ions for every formula unit dissolved, so the concentration of Cl\u2212<\/sup>(aq) would be 2 M, not 1 M.\r\n\r\nIn addition, the total ion concentration is the sum of the individual ion concentrations. Thus for the 1 M NaCl, the total ion concentration is 2 M; for the 1 M CaCl2<\/sub>, the total ion concentration is 3 M.\r\n

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

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