Learning Objectives Express the amount of solute in a solution in various concentration units. To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms with meanings that depend on various factors. Introduction Concentration is the measure of how much of a given substance is mixed with another substance. Solutions are said to be either dilute or concentrated. When we say that vinegar is \(5\%\) acetic acid in water, we are giving the concentration. If we said the mixture was \(10\%\) acetic acid, this would be more concentrated than the vinegar solution. A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. To make juice, you have to mix the frozen juice concentrate from inside these containers with three or four times the container size full of water. Therefore, you are diluting the concentrated juice. In terms of solute and solvent, the concentrated solution has a lot of solute versus the dilute solution that would have a smaller amount of solute. The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration. There are many ways to express the concentration of a solution. In this section, we will consider mass percent, volume percent, mass-volume percent, and parts per. In the next section, we will look at molarity. Mass Percent There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: \[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber \] mass of solution = mass of solute + mass solvent If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100.0 \: \text{g}\) of water. The mass of the solution is mass of solution = 25.0g sugar + 100.0g water = 125.0 g The percent by mass would be calculated by: \[\text{Percent by mass} = \dfrac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar} \nonumber \] Example \(\PageIndex{1}\) 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? Solution We can substitute the quantities given in the equation for mass/mass percent: \(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\) Exercise \(\PageIndex{1}\) A dextrose (also called D-glucose, C 6 H 12 O 6 ) solution with a mass of 2.00 × 10 2 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? Answer 7.90 % Using Mass Percent in Calculations Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form \(\dfrac{g \; \rm{solute}}{100 \; \rm{g solution}}\) or \(\dfrac{100 \; \rm g solution}{g\; \rm{solute}}\) For example, if you need to make \(3000.0 \: \text{g}\) of a \(5.00\%\) solution of sodium chloride, the mass of solute needs to be determined. Solution Given: 3000.0 g NaCl solution 5.00% NaCl solution Find: mass of solute = ? g NaCl Other known quantities: 5.00 g NaCl is to 100 g solution The appropriate conversion factor (based on the given mass percent) can be used follows: To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor. \[g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \dfrac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl \nonumber \] You would need to weigh out \(150 \: \text{g}\) of \(\ce{NaCl}\) and add it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{g} \right)\) from the mass of solution \(\left( 3000 \: \text{g} \right)\) to calculate the mass of the water that would need to be added. Exercise \(\PageIndex{1}\) What is the amount (in g) of hydrogen peroxide (H 2 O 2 ) needed to make a 6.00 kg, 3.00 % (by mass) H 2 O 2 solution? Answer 180 g H 2 O 2 Volume Percentage Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage , %vol or (v/v)%: \[ \text{volume percentage} = \dfrac{\text{volume solute}}{\text{volume solution}} \times100\%] Example \(\PageIndex{3}\): Calculations using Volume Percentage Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol? Solution Per the definition of volume percentage, the isopropanol volume is 70% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass: \[ \text {355 mL solution}(\frac{\text{70 mL isopropyl alcohol}}{\text{100 mL solution}})(\frac{\text{0.785 g isopropyl alcohol}}{\text{1 mL isopropyl alcohol}})=\text{195 g isopropyl alcohol} \nonumber \] Exercise \(\PageIndex{3}\) Wine is approximately 12% ethanol (\(\ce{CH_3CH_2OH}\)) by volume. Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL. How many moles of ethanol are present in a 750-mL bottle of wine? Answer 1.5 mol ethanol Mass-Volume Percentage “Mixed” percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of 0.9% mass/volume (m/v), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as “blood sugar”) is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter (100 mL) of blood (Figure \(\PageIndex{2}\)). Parts per Million and Parts per Billion Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb) . Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules. The mass-based definitions of ppm and ppb are given here: \[\text{ppm}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^6\; \text{ppm} \label{3.5.3A} \] \[\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^9\; \text{ppb} \label{3.5.3B} \] Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure \(\PageIndex{3}\)). Example \(\PageIndex{4}\): Parts per Million and Parts per Billion Concentrations According to the EPA, when the concentration of lead in tap water reaches 15 ppb, certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead (μg) would be contained in a typical glass of water (300 mL)? Solution The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm. Comparing these two unit definitions shows that ppm is 1000 times greater than ppb (1 ppm = 10 3 ppb). Thus: \[ \mathrm{15\; \cancel{ppb} \times \dfrac{1\; ppm}{10^3\;\cancel{ppb}} =0.015\; ppm} \nonumber \] The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields: \[\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} ×10^9\; \text{ppb} \nonumber \] \[\text{mass solute} = \dfrac{\text{ppb} \times \text{mass solution}}{10^9\;\text{ppb}} \nonumber \] \[\text{mass solute}=\mathrm{\dfrac{15\:ppb×300\:mL×\dfrac{1.00\:g}{mL}}{10^9\:ppb}=4.5 \times 10^{-6}\;g} \nonumber \] Finally, convert this mass to the requested unit of micrograms: \[\mathrm{4.5 \times 10^{−6}\;g \times \dfrac{1\; \mu g}{10^{−6}\;g} =4.5\; \mu g} \nonumber \] Exercise \(\PageIndex{4}\) A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units. Answer 9.6 ppm, 9600 ppb Measures of Concentration: https://youtu.be/RjMGaUpkg8g Summary In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.