Expressing Concentrations of Solutions

A complete description of a solution states what the solute is and how much solute is dissolved in a given amount of solvent or solution. The quantitative relationship between solute and solvent is the concentration of the solution. This concentration may be expressed using several different methods, as discussed next.

A. Concentration by Mass
The concentration of a solution may be given as the mass of solute in a given amount of solution, as in the following statements: The northern part of the Pacific Ocean contains 35.9 g salt in each 1000 g seawater. The North Atlantic Ocean has a higher salt concentration, 37.9 g salt/1000 g seawater.

B. Concentration by Percent
The concentration of a solution is often expressed as percent concentration by mass or percent by volume of solute in solution. Percent by mass is calculated from the mass of solute in a given mass of solution. A 5%-by-mass aqueous solution of sodium chloride contains 5 g sodium chloride and 95 g water in each 100 g solution.

 Percent by mass = mass of solutemass of solution X 100%

 Example: How many grams of glucose and of water are in 500 g of a 5.3% by-mass glucose solution Solution We know that 5.3% of the solution is glucose: The remainder of the 500 g is water:

If both solute and solvent are liquids, the concentration may be expressed as percent by volume. Both ethyl alcohol and water are liquids; the concentration of alcohol-water solutions is often given as percent by volume. For example, a 95% solution of ethyl alcohol contains 95 mL ethyl alcohol in each 100 mL solution.

 Percent by volume = volume of solutevolume of solution X 100%

 Example: Rubbing alcohol is an aqueous solution containing 70% isopropyl alcohol by volume. How would you prepare 250 mL rubbing alcohol from pure isopropyl alcohol? Solution We know that 70% of the volume is isopropyl alcohol: To prepare the solution, enough water is added to 175 mL isopropyl alcohol to form 250 mL solution.:

Because the density of liquids changes slightly as the temperature changes, a concentration given in percent by mass is accurate over a wider range of temperatures than is a concentration given in percent by volume. Sometimes a combination of mass and volume is used to express the concentration--the mass of solute dissolved in each 100 mL solution. Using this method, a 5% (wt/vol) solution of sodium chloride contains 5 g sodium chloride in each 100 mL solution.

C. Concentration in Parts per Million (ppm) and Parts per Billion (ppb) The terms (ppm) and parts per billion (ppb) are encountered more and more frequently as we become aware of the effects of substances present in trace amounts in water and air, and as we develop instruments sensitive enough to detect substances present in such low concentrations. In discussing mass, parts per million means concentration in grams per 106 grams, or micrograms per gram. In discussing volume, parts per million may mean milliliters per cubic meter, or the mixed designation of milligrams per cubic meter. For parts per billion, the general trend is toward the use of micrograms per liter when discussing water contaminants, micrograms per cubic meter for air, and micrograms per kilogram for soil concentrations.

D. Concentration in Terms of Moles
The concentration of a solution may be stated as molarity (M), which is the number of moles of solute per liter of solution or the number of millimoles (mmol) (1 millimole = 10-3 mole) per milliliter of solution.

 Molarity (M) = moles solutevolume (liter) solution = millimoles solutemilliliter solution

A 6 M (say "six molar") solution of hydrochloric acid contains 6 mol hydrochloric acid in 1 L solution.

The molarity of a solution gives a ratio between moles of solute and volume of solution. It can be used as a conversion factor between these two units in calculations involving solutions. As a conversion factor, it can be used two ways:

1. Moles/volume (L) states the number of moles in one liter of solution. This conversion factor is used in calculating the number of moles of solute in a given volume of solution.

2. Volume (L)/moles states that one liter contains some number of moles of solution. This conversion factor is used to calculate the volume of a solution that contains a given quantity of solute.

 Example:How many moles of hydrochloric acid are in 200 mL of 0.15 M HCl? Solution Wanted: ? mol HCl Given 200 mL of 0.15 M HCl Conversion Factors Equation Answer 0.30 mol HCl Note that, each time the volume of a solution is stated, the concentration of the solution is given. This form may look confusing, but without this marking it is easy to forget which solution you are referring to.

 Example:What mass of sodium hydroxide (NaOH) is needed to prepare 100 ML of 0.125 M sodium hydroxide? Solution Wanted: ? g NaOH Given 100 mL of 0.125 M NaOH Conversion Factors 1 L of 0.125 M NaOH contains 0.125 mol NaOH- that is, Equation Answer 0.500 g NaOH

 Example: What volume of 3.25 M sulfuric acid is needed to prepare 0.500 L of 0.130 M H2SO4? Solution We are to prepare 0.500 L of 0.130 M H2SO4 by adding an amount of water to an amount of 3.25 M H2SO4. The moles of sulfuric acid in the final (more dilute) solution will be the same as the moles of sulfuric acid in the portion of the more concentrated solution. We can calculate the moles of sulfuric acid in the final dilute solution: This answer gives the moles of acid needed. We can calculate the volume of 3.25 M H2SO4 that would contain 0.065 mol H2SO4. This answer gives the volume of concentrated acid that conatins the moles of acid needed for the dilute solution. This volume of 3.25 M H2SO4 would be dissolved in 480 mL (500 mL - 20 mL) water to prepare 0.500 L of 0.130 M H2SO4. This problem is diagramed in the figure.

 Example: What volume of 6.39 M sodium chloride contains 51.2 mmol sodium chloride? Solution Wanted ? mL of 6.39 M NaCl Given 51.2 mmol NaCl Conversion factors 1 L of 6.39 M NaCl contains 6.39 mol NaCl 1 mL of 6.39 M NaCl contains 6.39 mmol NaCl Equation Answer 8.01 mL of 6.39 M NaCl

 Example: How do we prepare 75.0 mL of 0.96 M sulfuric acid from 18 M acid? Solution We are to prepare 75.0 mL of 0.96 M sulfuric aicd by diluting 18 M sulfuric acid with water. We can calculate the millimoles of sulfuric acid in the final solution: We can calculate the volume of 18 M ? mmol H2SO4 that will contain 72 mmol H2SO4: The solution would be prepared by adding 4.0 mL of 18 M H2SO4 to about 50 mL of water and then dilutint that solution to exactly 75.0 mL.

Table 11.3 lists several of the commonly used ways of expressing concentrations.

  Solute   Solvent   Solution Percent by weight ? g + ? g 100 g accurate, independent of temperature Percent by volume ? mL + ? mL 100 mL used when solute is liquid; concentration varies slightly with tenperature Percent, weight/volume ? g + 100 mL used when solute is liquid; concentration varies slightly with tenperature Molarity (M) moles 1 liter } used in chemical calculations Millimole/liter 10-3 mole 1 liter Millimole/milliliter 10-3 mole 10-3 liter Parts per million (ppm) mg kg } used in environmental studies Parts per billion (ppb) µg kg