{"id":814,"date":"2018-03-20T16:08:52","date_gmt":"2018-03-20T16:08:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=814"},"modified":"2018-09-25T20:31:19","modified_gmt":"2018-09-25T20:31:19","slug":"9-2-concentration","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-monroecc-orgbiochemistry\/chapter\/9-2-concentration\/","title":{"raw":"9.2 Concentration","rendered":"9.2 Concentration"},"content":{"raw":"
\r\n
<\/div>\r\n
9.2<\/span> Concentration<\/span><\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

Learning Objectives<\/h3>\r\n
    \r\n \t
  1. Express the amount of solute in a solution in various concentration units.<\/li>\r\n \t
  2. Use molarity to determine quantities in chemical reactions.<\/li>\r\n \t
  3. Determine the resulting concentration of a diluted solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

    To define a solution precisely, we need to state its concentration<\/span><\/span>: how much solute is dissolved in a certain amount of solvent. Words such as dilute<\/em> or concentrated<\/em> are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors.<\/p>\r\n\r\n

    \r\n

    Solubility<\/h2>\r\n

    There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility<\/span><\/span><\/strong>\u00a0of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table 9.2 \"Solubilities of Various Solutes in Water at 25\u00b0C \"<\/a> lists the solubilities of various solutes in water. Solubility varies with temperature, so it is important to include the temperature at which the solubility was determined.\u00a0 Most solids have a higher solubility in water at higher temperatures. For example, while 91 g of glucose will dissolve in 100 mL of water at 25o<\/sup>C, 357 g will dissolve at 70o<\/sup>C.<\/p>\r\n\r\n

    \r\n

    Table 9.2<\/span> Solubilities of Various Solutes in Water at 25\u00b0C<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Substance<\/th>\r\nSolubility (g in 100 mL of H2<\/sub>O)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    AgCl(s)<\/td>\r\n0.019<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H6<\/sub>(\u2113) (benzene)<\/td>\r\n0.178<\/td>\r\n<\/tr>\r\n
    CH4<\/sub>(g)<\/td>\r\n0.0023<\/td>\r\n<\/tr>\r\n
    CO2<\/sub>(g)<\/td>\r\n0.150<\/td>\r\n<\/tr>\r\n
    CaCO3<\/sub>(s)<\/td>\r\n0.058<\/td>\r\n<\/tr>\r\n
    CaF2<\/sub>(s)<\/td>\r\n0.0016<\/td>\r\n<\/tr>\r\n
    Ca(NO3<\/sub>)2<\/sub>(s)<\/td>\r\n143.9<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H12<\/sub>O6<\/sub> (glucose)<\/td>\r\n90.9<\/td>\r\n<\/tr>\r\n
    KBr(s)<\/td>\r\n67.8<\/td>\r\n<\/tr>\r\n
    MgCO3<\/sub>(s)<\/td>\r\n2.20<\/td>\r\n<\/tr>\r\n
    NaCl(s)<\/td>\r\n36.0<\/td>\r\n<\/tr>\r\n
    NaHCO3<\/sub>(s)<\/td>\r\n8.41<\/td>\r\n<\/tr>\r\n
    C12<\/sub>H22<\/sub>O11<\/sub> (sucrose)<\/td>\r\n211.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

    If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated<\/strong>. <\/span><\/span>If a solution contains less solute than the solubility limit, it is unsaturated<\/span><\/strong><\/span>. For some substances, a\u00a0supersaturated <\/strong>solution can be prepared by dissolving as much solute as possible at a higher temperature, then lowering the temperature.\u00a0 Less solute should dissolve at the lower temperature, but the excess solute might not precipitate right away.\u00a0 Thus the solution is supersaturated, but not stable. Eventually the excess solute will crystallize with a release of energy as the mixture achieves a more stable state.\u00a0 Honey is a supersaturated solution which bees form by using their wings to fan nectar, a dilute sugar solution, to drive off much of the water, forming supersaturated honey.\u00a0 The excess sugar in honey sometimes recrystallizes, making the honey cloudy or grainy.\u00a0 It can be restored to a clear state by heating it.<\/p>\r\n

    In contrast to increased solubility of most solids at higher temperatures, the solubility of gases in liquid water decreases as temperature increases.\u00a0 The solubility of O2<\/sub> is 8.2 mg O2<\/sub>\/L of H2<\/sub>O at 25o<\/sup>C but only 6.0\u00a0mg O2<\/sub>\/L of H2<\/sub>O at 45o<\/sup>C, both at a pressure of 1 atm. \u00a0\u00a0 Pressure must be specified because it also influences the solubility of gases; more gas dissolves when the pressure is higher. \u00a0 At 0.5 atm, 4.1 mg of O2<\/sub> dissolves per L of H2<\/sub>O, but at 2 atm, the solubility of O2<\/sub> is 16.5 mg\/L, both at a temperature of 25\u00a0o<\/sup>C. \u00a0 When pressure is suddenly changed, such as when you open a can of soda, excess gas can bubble out of solution uncontrollably.<\/p>\r\n\r\n

    \r\n
    \r\n
    \r\n
    \r\n

    Note<\/h3>\r\n

    Reusable hot packs contain a supersaturated solution of sodium acetate.\u00a0 When recrystallization is initiated using a clicker, the excess sodium acetate (NaC2<\/sub>H3<\/sub>O2<\/sub>) comes out of solution and heat is emitted.\u00a0 The hot pack can be made ready for reuse by placing it in a pan of boiling water until the sodium acetate dissolves.<\/p>\r\n\r\n<\/div>\r\n

    Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution.\u00a0<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    Percent Composition<\/h2>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    There are several ways of expressing the concentration of a solution by using a percentage. The mass\/mass percent<\/span><\/span>\u00a0(% m\/m) is defined as the mass of a solute divided by the mass of a solution times 100:<\/p>\r\n[latex]\\text{%m\/m}=\\frac{\\text{mass solute}}{\\text{mass solution}}\\times{100}[\/latex]\r\n

    Each mass must be expressed in the same units to determine the proper concentration.\u00a0 The mass of the solution may be given, or it may be necessary to calculate it by adding the masses of the solute and solvent.<\/p>\r\n\r\n

    \r\n

    Example 2<\/h3>\r\n

    A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass\/mass percent concentration of the solution?<\/p>\r\n

    Solution<\/p>\r\n

    We can substitute the quantities given in the equation for mass\/mass percent:<\/p>\r\n[latex]\\text{%m\/m}=\\frac{\\text{mass solute}}{\\text{mass solution}}\\times{100}=\\frac{36.5\\text{ g}}{355\\text{ g}}\\times{100}=10.3\\text{%}[\/latex]\r\n\r\n<\/div>\r\n

    \r\n
    \r\n

    Skill-Building Exercise<\/h3>\r\n
      \r\n \t
    1. \r\n
      \r\n

      A dextrose (also called D-glucose, C6<\/sub>H12<\/sub>O6<\/sub>) solution with a mass of 2.00 \u00d7 102<\/sup> g has 15.8 g of dextrose dissolved in it. What is the mass\/mass percent concentration of the solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

      For liquids, volumes are relatively easy to measure, so the concentration of a liquid in solution can be expressed as a volume\/volume percent<\/span><\/span>\u00a0(% v\/v): the volume of a solute divided by the volume of a solution times 100:<\/p>\r\n[latex]\\text{%v\/v}=\\frac{\\text{volume solute}}{\\text{volume solution}}\\times{100}[\/latex]\r\n

      As with mass\/mass %, the units of the solute and the solution must be the same.\u00a0 Both the volume of the solute and the volume of the solution must bespecified because volumes are not additive.\u00a0 For example, a mixture made from 50 mL of water and 50 mL of ethanol has a volume of about 95 ml because the molecules of the two substances are attracted to each other and enter the spaces between each other's molecules.<\/p>\r\n

      A hybrid concentration unit, mass\/volume percent<\/span><\/span>\u00a0(% m\/v), is commonly used for intravenous (IV) fluids (Figure 9.2 \"Mass\/Volume Percent\"<\/a>). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:<\/p>\r\n[latex]\\text{%m\/v}=\\frac{\\text{grams solute}}{\\text{mL solution}}\\times{100}[\/latex]\r\n\r\nUnlike %m\/m and %m\/v, the units do not cancel and m\/v% is not a true percentage.\u00a0 Never the less, m\/v % is widely used in medical settings because it is easy to deliver a certain mass of a medication or other solute by delivering a certain volume of solution.\r\n<\/span>\r\n

      Each percent concentration can be used to produce a conversion factor between the amount of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.\u00a0 While it is possible to substitute known quantities into the percentage equation and solve for the missing quantity, it is easier to interpret percentage and use it as a conversion factor.\u00a0 x % implies the following pair of conversion factors:<\/p>\r\n[latex]\\frac{\\text{x g or mL of solute}}{100\\text{ g or mL of solution}}[\/latex]\u00a0\u00a0 AND\u00a0\u00a0 [latex]\\frac{100\\text{ g or mL of solution}}{\\text{ x g or mL of solute}}[\/latex]\u00a0\u00a0\u00a0 As always, use the conversion factor that allows the units to cancel.\u00a0 Make sure to include a descriptor with the units, such as mL solution or g solvent- cancellation must include the unit and the descriptor.\u00a0 Note that the 100 is placed there by definition of percent, so it does not limit the sig figs of the answer.\r\n

      \r\n

      Example 3<\/h3>\r\n

      A sample of 45.0% v\/v solution of ethanol (C2<\/sub>H5<\/sub>OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?<\/p>\r\n

      Solution<\/p>\r\n

      There are two pieces of information given, the volume and the percent.\u00a0 When percent is given, it will be used as the conversion factor, so begin with the volume.<\/p>\r\n[latex]115\\cancel{\\text{ mL solution}}\\times{\\frac{45.0\\text{ mL ethanol}}{100\\cancel{\\text{ mL solution}}}}=51.8\\text{ mL ethanol}[\/latex]\r\n\r\nThe answer makes sense because a bit more than 100 mL of solution should contain a bit more than 45 g of ethanol.\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      Note<\/h3>\r\n

      The highest concentration of ethanol that can be obtained normally is 95% ethanol, which is actually 95% v\/v.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      Skill-Building Exercise<\/h3>\r\n
        \r\n \t
      1. \r\n
        \r\n

        What volume of a 12.75% m\/v solution of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) in water is needed to obtain 50.0 g of C6<\/sub>H12<\/sub>O6<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

        \r\n
        \r\n

        Skill-Building Exercise<\/h3>\r\n
          \r\n \t
        1. \r\n
          \r\n

          The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass\/volume percent of the bleach?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

          In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm)\u00a0<\/span><\/span>and parts per billion (ppb)<\/span><\/span>. Both of these units can be based on either mass or volume and are defined as follows:<\/p>\r\n[latex]\\text{ppm}=\\frac{\\text{g or mL solute}}{\\text{g or mL solution}}\\times1,000,000[\/latex]\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0 [latex]\\text{ppb}=\\frac{\\text{g or mL solute}}{\\text{g or mL solution}}\\times1,000,000,000[\/latex]\r\n

          \r\n
          \r\n

          Note<\/h3>\r\n

          Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

          Concentrations of trace elements<\/em> in the body\u2014elements that are present in extremely low concentrations but are nonetheless necessary for life\u2014are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon\u2019s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.<\/p>\r\n\r\n

          \r\n
          \r\n

          Note<\/h3>\r\n

          In aqueous solutions, 1 ppm is essentially equal to 1 mg\/L, and 1 ppb is equivalent to 1 \u00b5g\/L.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

          \r\n

          Example 4<\/h3>\r\n

          If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?<\/p>\r\n

          Solution<\/p>\r\nThe starting point is the simpler value given, the body mass of 70.0 kg.\r\n\r\nA concentration of 21 ppb means \u201c21 g of solute per 1,000,000,000 g (or 109<\/sup>g)\u00a0 of body mass.\u201d\u00a0\u00a0 Written as a pair of conversion factors::\r\n\r\n[latex]\\frac{21\\text{ g Co}}{10^9\\text{ g body mass}}[\/latex]\u00a0\u00a0 AND\u00a0\u00a0 [latex]\\frac{19^9\\text{ g body mass}}{21\\text{ g Co}}[\/latex]\r\n\r\nBut before applying one of these conversion factors, the body mass must be converted from kg to g:\r\n

          [latex]70.0\\cancel{\\text{ kg}}\\times{\\frac{1000\\text{ g}}{1\\cancel{\\text{kg}}}}=7.00\\times{10^4}\\text{ g}[\/latex]<\/p>\r\n

          Now we determine the amount of Co:<\/p>\r\n[latex]7.00\\times{10^4}\\cancel{\\text{ g body mass}}\\times{\\frac{21\\text{ g Co}}{10^9\\cancel{\\text{ g body mass}}}}=0.0015\\text{ g Co}[\/latex]\r\n

          This is only 1.5 mg.<\/p>\r\n\r\n<\/div>\r\n

          \r\n
          \r\n

          Skill-Building Exercise<\/h3>\r\n
            \r\n \t
          1. \r\n
            \r\n

            An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

            \r\n

            Molarity<\/h2>\r\n

            Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions that take place in solution. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.<\/p>\r\n

            Molarity<\/span><\/span>\u00a0is defined as the number of moles of a solute dissolved per liter of solution:<\/p>\r\n[latex]\\text{molarity}=\\frac{\\text{number of moles solute}}{\\text{number of liters solution}}[\/latex]\r\n

            Notice that the units are complex, mol\/L, often abbreviated to just M.\u00a0 Also note that this is not<\/span> a percentage so there is no multiplication by 100.<\/p>\r\n

            For example, for 1.5 mol of NaCl dissolved in 0.500 L of solution, calculate the molarity as follows:<\/p>\r\n[latex]\\text{molarity}=\\frac{\\text{number of moles solute}}{\\text{number of liters solution}}=\\frac{1.5\\text{ mol NaCl}}{0.500\\text{ L solution}}=3.0\\text{ mol\/L}=3.0\\text{ M NaCl}[\/latex]\r\n

            Before<\/em> a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.<\/p>\r\n\r\n

            \r\n

            Example 5<\/h3>\r\n

            What is the molarity of an aqueous solution of 25.0 g of NaOH in 750. mL?<\/p>\r\n

            Solution<\/p>\r\n

            Before substituting these quantities into the definition of molarity, convert them to the proper units. The mass of NaOH must be converted to moles of NaOH using\u00a0 molar mass of NaOH as a conversion factor.\u00a0 The molar mass of NaOH is 40.00 g\/mol.<\/p>\r\n[latex]25.0\\cancel{\\text{ g NaOH}}\\times{\\frac{1 \\text{ mol NaOH}}{40.00\\cancel{\\text{g NaOH}}}}=0.625\\text{ mol NaOH}[\/latex]\r\n

            Next, convert the volume units from milliliters to liters:<\/p>\r\n[latex]750.\\cancel{\\text{ mL}}\\times{\\frac{1\\text{ L}}{1000\\cancel{\\text{ mL}}}}=0.750\\text{ L}[\/latex]\r\n

            Now that the quantities are expressed in the proper units, substitute them into the definition of molarity:<\/p>\r\n[latex]\\text{molarity}=\\frac{\\text{number of moles solute}}{\\text{number of liters solution}}=\\frac{0.625\\text{ mol NaOH}}{0.750\\text{ L solution}}=3.0\\text{ mol\/L}=0.833\\text{ M NaOH}[\/latex]\r\n\r\n<\/div>\r\n

            \r\n
            \r\n

            Skill-Building Exercise<\/h3>\r\n
              \r\n \t
            1. \r\n
              \r\n

              If a 350 mL cup of coffee contains 0.150 g of caffeine (C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>), what is the molarity of this caffeine solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

              When molarity is given, it can be interpreted as a pair of conversion factors that can be used to convert between moles of solute and liters of solution.\u00a0 A molariy of x :<\/p>\r\n[latex]\\frac{\\text{x moles solute}}{1\\text{ L solution}}[\/latex]\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0 [latex]\\frac{1\\text{ L solution}}{\\text{x moles solute}}[\/latex]\r\n\r\nNote that the 1 L is part of the definition of molarity, so it does not limit the number of sig figs in the answer.\r\n

              \r\n

              Example 6<\/h3>\r\nWhat volume of a 0.0753 M solution of dimethylamine [(CH3<\/sub>)2<\/sub>NH] is needed to obtain 0.450 mol of the compound?\r\n

              Solution<\/p>\r\n

              Begin with the given that has simpler units, 0.450 mol.\u00a0 Remember that M means mols\/L, a complex unit, so it will be used to create a conversion factor.<\/p>\r\n

              [latex]0.450\\text{ mol dimethylamine}\\times{\\frac{1\\text{ L solution}}{0.0753\\text{ moles dimethylamine}}}=5.98\\text{ L solution}[\/latex]<\/span><\/p>\r\n\r\n

              Example 7<\/h3>\r\nEthylene glycol (C2<\/sub>H6<\/sub>O2<\/sub>) is mixed with water to make auto engine coolants. How many grams of C2<\/sub>H6<\/sub>O2<\/sub> are in 5.00 L of a 6.00 M aqueous solution?\u00a0 The molar mass of C2<\/sub>H6<\/sub>O2<\/sub> is 62.08 g\/mol.\r\n

              Three bits of information are provided, molarity and molar mass which both have complex units and are likely to be used as conversion factors, and volume which has a simple unit and thus is the starting point.<\/p>\r\n[latex]5.00\\cancel{\\text{ L solution}}\\times{\\frac{6.00\\cancel{\\text{ mol ethylene glycol}}}{1\\cancel{\\text{ L solution}}}}\\times{\\frac{62.08\\text{ g ethylene glycol}}{1\\cancel{\\text{ mol ethylene glycol}}}}=1860\\text{ g ethylene glycol}[\/latex]\r\n\r\n<\/div>\r\n

              \r\n
              \r\n
              \r\n

              Note<\/h3>\r\n

              Dimethylamine has a \u201cfishy\u201d odor. In fact, organic compounds called amines cause the odor of decaying fish. (For more information about amines, see Chapter 15 \"Organic Acids and Bases and Some of Their Derivatives\"<\/a>, Section 15.1 \"Functional Groups of the Carboxylic Acids and Their Derivatives\"<\/a> and Section 15.11 \"Amines: Structures and Names\"<\/a> through Section 15.13 \"Amines as Bases\"<\/a>.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

              \r\n
              \r\n

              Skill-Building Exercise<\/h3>\r\n<\/div>\r\n
              \r\n
                \r\n \t
              1. \r\n
                \r\n

                What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              2. \r\n
                \r\n

                Acetic acid (HC2<\/sub>H3<\/sub>O2<\/sub>) is the acid in vinegar. How many grams of HC2<\/sub>H3<\/sub>O2<\/sub> are in 0.565 L of a 0.955 M solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

                \r\n

                Using Molarity in Stoichiometry Problems<\/h2>\r\n

                Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit.<\/p>\r\n

                The general steps for performing stoichiometry problems using molarities of solutions\u00a0 are shown in Figure 9.3 \"Diagram of Steps for Using Molarity in Stoichiometry Calculations\"<\/a>. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure 9.3 \"Diagram of Steps for Using Molarity in Stoichiometry Calculations\"<\/a> indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.<\/p>\r\n\r\n

                \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"506\"]\"image\" Figure 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations <\/em>[\/caption]\r\n\r\n<\/div>\r\n

                In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.<\/p>\r\n\r\n

                \r\n

                Example 8<\/h3>\r\n

                What volume of a 2.75 M HCl solution is needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:<\/p>\r\nHCl(aq) + NaOH(s) \u2192 H2<\/sub>O(\u2113) + NaCl(aq)<\/span><\/span>\r\n

                Solution<\/p>\r\n

                This problem gives grams of NaOH, so converting to moles NaOH using NaOH's molar mass is the first step.\u00a0 Next is the central part of any stoichiometry problem, relating moles of one substance to moles of a different substance using the mole-to-mole ratio based on coefficients from the balanced equation. Finally, the volume of HCl solution is the unknown, so the molarity of the HCl solution becomes the conversion factor between g HCl and volume of HCl solution.<\/p>\r\n[latex]185\\cancel{\\text{ g NaOH}}\\times{\\frac{1\\cancel{\\text{ mol NaOH}}}{40.00\\cancel{\\text{ g NaOH}}}}\\times{\\frac{1\\cancel{\\text{ mol HCl}}}{1\\cancel{\\text{ mol NaOH}}}}\\times{\\frac{1\\text{ L HCl solution}}{2.75\\cancel{\\text{mol HCl}}}}=1.68\\text{ L HCl solution}[\/latex]\r\n\r\nThe answer could also be expressed in mL by multiplying by 1000, so 1680 mL.\r\n\r\n<\/div>\r\n

                \r\n
                \r\n

                Skill-Building Exercise<\/h3>\r\n
                  \r\n \t
                1. \r\n
                  \r\n

                  How many milliliters of a 1.04 M H2<\/sub>SO4<\/sub> solution are needed to react with 98.5 g of Ca(OH)2<\/sub>? The balanced chemical equation for the reaction is as follows:<\/p>\r\nH2<\/sub>SO4<\/sub>(aq) + Ca(OH)2<\/sub>(s) \u2192 2H2<\/sub>O(\u2113) + CaSO4<\/sub>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                  \r\n

                  Concentrations of Substances in Bodily Fluids<\/h2>\r\n<\/div>\r\n

                  Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table 9.3 \"Approximate Concentrations of Various Solutes in Some Solutions in the Body*\"<\/a> lists the typical concentrations of some of these solutes.<\/p>\r\n\r\n

                  \r\n

                  Table 9.3<\/span> Approximate Concentrations of Various Solutes in Some Solutions in the Body*<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                  Solution<\/th>\r\nSolute<\/th>\r\nConcentration (M)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                  blood plasma<\/td>\r\nNa+<\/sup><\/td>\r\n0.138<\/td>\r\n<\/tr>\r\n
                  K+<\/sup><\/td>\r\n0.005<\/td>\r\n<\/tr>\r\n
                  Ca2+<\/sup><\/td>\r\n0.004<\/td>\r\n<\/tr>\r\n
                  Mg2+<\/sup><\/td>\r\n0.003<\/td>\r\n<\/tr>\r\n
                  Cl\u2212<\/sup><\/td>\r\n0.110<\/td>\r\n<\/tr>\r\n
                  HCO3<\/sub>\u2212<\/sup><\/td>\r\n0.030<\/td>\r\n<\/tr>\r\n
                  stomach acid<\/td>\r\nHCl<\/td>\r\n0.10<\/td>\r\n<\/tr>\r\n
                  urine<\/td>\r\nNaCl<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
                  PO4<\/sub>3\u2212<\/sup><\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n
                  NH2<\/sub>CONH2<\/sub> (urea)<\/td>\r\n0.30<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n
                  *Note: Concentrations are approximate and can vary widely.<\/th>\r\n<\/tr>\r\n<\/tfoot>\r\n<\/table>\r\n<\/div>\r\n
                  \r\n
                  \r\n
                  \r\n
                  \r\n

                  Looking Closer: The Dose Makes the Poison<\/h3>\r\n

                  Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison<\/em>. This means that what may be dangerous in some amounts may not be dangerous in other amounts.<\/p>\r\n

                  Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.<\/p>\r\n

                  Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

                  Expressing Concentration of Ionic Substances in Equivalents<\/h2>\r\n<\/div>\r\n<\/div>\r\n
                  \r\n

                  Concentrations of ionic solutes are occasionally expressed in units called equivalents<\/strong> (Eq)<\/span><\/span>. One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol\/L of Na+<\/sup>(aq) is also 1 Eq\/L because sodium has a 1+ charge. A 1 mol\/L solution of Ca2+<\/sup>(aq) ions has a concentration of 2 Eq\/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)\u2014for example, human blood plasma has a total concentration of about 150 mEq\/L. (For more information about the ions present in blood plasma, see Chapter 3 \"Ionic Bonding and Simple Ionic Compounds\"<\/a>, Section 3.3 \"Formulas for Ionic Compounds\"<\/a>.)<\/p>\r\n\r\n<\/div>\r\n

                  \r\n

                  Dilution<\/h2>\r\n

                  When solvent is added to dilute a solution, the volume of the solution changes, but the amount of solute does not change. Before dilution, the amount of solute was equal to its original concentration (Ci<\/sub>) times its original volume (Vi<\/sub>).<\/p>\r\n

                  amount of substance = Ci<\/sub> \u00d7 Vi<\/sub><\/span><\/span><\/p>\r\n

                  After dilution, the same amount of solute is equal to the final concentration\u00a0(Cf<\/sub>) times the final volume (Vf<\/sub>):<\/p>\r\n

                  amount of substance = Cf<\/sub> \u00d7 Vf<\/sub><\/span><\/span><\/p>\r\n

                  Since the amount of substance does not change during dilution:<\/p>\r\n

                  Ci<\/sub> \u00d7 Vi<\/sub>= Cf<\/sub> \u00d7 Vf<\/sub><\/span><\/span><\/p>\r\n

                  Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit.\u00a0 Three of the values must be given in the problem so that the fourth value can be determined algebraically.<\/p>\r\n\r\n

                  \r\n

                  Example 9<\/h3>\r\n

                  A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?<\/p>\r\n

                  Solution<\/p>\r\nThe numbers given in the problem are Vi<\/sub>, Ci<\/sub>, and Vf<\/sub>, respectively.\u00a0 Cf<\/sub> is the unknown.\u00a0 To solve for Cf<\/sub>, divide both sides by Vf<\/sub>.\r\n

                  [latex]\\frac{\\text{C}_\\text{i}\\times{\\text{V}_\\text{i}}}{\\text{V}_\\text{f}}=\\frac{\\text{C}_\\text{f}\\times{\\text{V}_\\text{f}}}{\\text{V}_\\text{f}}[\/latex]\u00a0\u00a0 so after cancelling\u00a0\u00a0 [latex]\\text{C}_\\text{f}=\\frac{\\text{C}_\\text{i}\\times{\\text{V}_\\text{i}}}{\\text{V}_\\text{f}}[\/latex]<\/p>\r\nFilling in the actual values, [latex]\\text{C}_\\text{f}=\\frac{0.900\\text{ M}\\times{125\\text{ mL}}}{1125\\text{M}}=0.100\\text{ M}[\/latex]\r\n\r\nThe answer is reasonable because dilution produces a larger volume with a smaller concentration.\r\n\r\n<\/div>\r\n

                  \r\n
                  \r\n

                  Skill-Building Exercise<\/h3>\r\n
                    \r\n \t
                  1. \r\n
                    \r\n

                    A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                    \r\n
                    \r\n

                    Concept Review Exercises<\/h3>\r\n
                      \r\n \t
                    1. \r\n
                      \r\n

                      What are some of the units used to express concentration?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                    2. \r\n
                      \r\n

                      Distinguish between the terms solubility<\/em> and concentration<\/em>.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

                      \r\n

                      Answers<\/h3>\r\n
                      <\/div>\r\n
                      \r\n
                        \r\n \t
                      1. \r\n
                        \r\n

                        % m\/m, % m\/v, ppm, ppb, molarity, and Eq\/L (answers will vary)<\/p>\r\n\r\n<\/div><\/li>\r\n \t

                      2. \r\n
                        \r\n

                        Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

                        \r\n

                        Key Takeaways<\/h3>\r\n
                        <\/div>\r\n
                        \r\n