2

Solutions

Solutions are homogeneous mixtures. Many solutions contain one component called the solvent in which other components called solutes are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution. Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water.

By the end of this section, you will be able to:

- Describe the fundamental properties of solutions
- Calculate solution concentrations using molarity
- Perform dilution calculations using the dilution equation

Preceding sections of this chapter focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 2.1). This section will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

Solutions have previously been defined as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. A more thorough treatment of solution properties is provided in the chapter on solutions and colloids, but provided here is an introduction to some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course very common on our planet. A solution in which water is the solvent is called an aqueous solution.

A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (*M*) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

$$M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}$$

$$M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{0.133\phantom{\rule{0.2em}{0ex}}\text{mol}}{355\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{L}}{1000\phantom{\rule{0.2em}{0ex}}\text{mL}}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}=0.375\phantom{\rule{0.2em}{0ex}}M$$

0.05 M

$$\begin{array}{c}\\ M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}\\ \text{mol solute}=M\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{L solution}\\ \\ \text{mol solute}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.375\phantom{\rule{0.4em}{0ex}}\frac{\text{mol sugar}}{\text{L}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\left(10\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{L}}{1000\phantom{\rule{0.2em}{0ex}}\text{mL}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.004\phantom{\rule{0.2em}{0ex}}\text{mol sugar}\end{array}$$

80 mL

$$M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\mathrm{25.2\; g}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}{\text{mol CH}}_{3}{\text{CO}}_{2}\text{H}}{{\text{60.052 g CH}}_{3}{\text{CO}}_{2}\text{H}}\phantom{\rule{0.2em}{0ex}}}{\text{0.500 L solution}}\phantom{\rule{0.2em}{0ex}}=0.839\phantom{\rule{0.2em}{0ex}}M$$

$$\begin{array}{l}\\ M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}=0.839\phantom{\rule{0.2em}{0ex}}M\\ M=\phantom{\rule{0.2em}{0ex}}\frac{0.839\phantom{\rule{0.2em}{0ex}}\text{mol solute}}{1.00\phantom{\rule{0.2em}{0ex}}\text{L solution}}\phantom{\rule{0.2em}{0ex}}\end{array}$$

Calculate the molarity of 6.52 g of CoCl_{2} (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

0.674 *M*

How many grams of NaCl are contained in 0.250 L of a 5.30-*M* solution?

The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 2.2:

$$\begin{array}{c}\\ M=\phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{L solution}}\phantom{\rule{0.2em}{0ex}}\\ \text{mol solute}=M\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{L solution}\\ \\ \text{mol solute}=5.30\phantom{\rule{0.2em}{0ex}}\frac{\text{mol NaCl}}{\text{L}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}0.250\phantom{\rule{0.2em}{0ex}}\text{L}=1.325\phantom{\rule{0.2em}{0ex}}\text{mol NaCl}\end{array}$$

Finally, this molar amount is used to derive the mass of NaCl:

$$\text{1.325 mol NaCl}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{58.44\phantom{\rule{0.2em}{0ex}}\text{g NaCl}}{\text{mol NaCl}}\phantom{\rule{0.2em}{0ex}}=77.4\phantom{\rule{0.2em}{0ex}}\text{g NaCl}$$

5.55 g CaCl_{2}

When performing calculations stepwise, as in Example 2.4, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 2.4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If the guard digit had not been retained, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, rounding errors may also be avoided by performing computations in a single step (see Example 2.5). This eliminates intermediate steps so that only the final result is rounded.

$$\text{g solute}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{g solute}}\phantom{\rule{0.2em}{0ex}}=\text{mol solute}$$

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

$$\text{mol solute}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{L solution}}{\text{mol solute}}\phantom{\rule{0.2em}{0ex}}=\text{L solution}$$

Combining these two steps into one yields:

$$\text{g solute}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{mol solute}}{\text{g solute}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{L solution}}{\text{mol solute}}\phantom{\rule{0.2em}{0ex}}=\text{L solution}$$

$$75.6\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\phantom{\rule{0.2em}{0ex}}\left(\frac{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\phantom{\rule{0.2em}{0ex}}}{60.05\phantom{\rule{0.2em}{0ex}}\text{g}}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{\text{L solution}}{0.839\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.50\phantom{\rule{0.2em}{0ex}}\text{L solution}$$

0.370 L

Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 2.3).

Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated *stock solution*, a solution of lesser concentration may be prepared. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents.

A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the number of moles of solute in a solution (*n*) is equal to the product of the solution’s molarity (*M*) and its volume in liters (*L*):

$$n=ML$$

Expressions like these may be written for a solution before and after it is diluted:

$${n}_{1}={M}_{1}{L}_{1}$$

$${n}_{2}={M}_{2}{L}_{2}$$

where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process *does not change the amount of solute in the solution, **n*_{1} = *n*_{2}. Thus, these two equations may be set equal to one another:

$${M}_{1}{L}_{1}={M}_{2}{L}_{2}$$

This relation is commonly referred to as the dilution equation. Although this equation uses molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used as long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:

$${C}_{1}{V}_{1}={C}_{2}{V}_{2}$$

where *C* and *V* are concentration and volume, respectively.

Use the simulation to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.

$$\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\ \\ {C}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{{C}_{1}{V}_{1}}{{V}_{2}}\phantom{\rule{0.2em}{0ex}}\end{array}$$

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), the diluted solution’s concentration is expected to be less than one-half 5 *M*. This ballpark estimate will be compared to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

$${C}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{0.850\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}5.00\phantom{\rule{0.2em}{0ex}}\frac{\text{mol}}{\text{L}}\phantom{\rule{0.2em}{0ex}}}{\mathrm{1.80\; L}}\phantom{\rule{0.2em}{0ex}}=2.36\phantom{\rule{0.2em}{0ex}}M$$

This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 *M*).

0.102 *M* CH_{3}OH

$$\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\ \\ {V}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{{C}_{1}{V}_{1}}{{C}_{2}}\phantom{\rule{0.2em}{0ex}}\end{array}$$

Since the diluted concentration (0.12 *M*) is slightly more than one-fourth the original concentration (0.45 *M*), the volume of the diluted solution is expected to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

$$\begin{array}{c}\\ {V}_{2}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.45\phantom{\rule{0.2em}{0ex}}M\right)\left(0.011\phantom{\rule{0.2em}{0ex}}\text{L}\right)}{\left(0.12\phantom{\rule{0.2em}{0ex}}M\right)}\phantom{\rule{0.2em}{0ex}}\\ {V}_{2}=0.041\phantom{\rule{0.2em}{0ex}}\text{L}\end{array}$$

The volume of the 0.12-*M* solution is 0.041 L (41 mL). The result is reasonable and compares well with the rough estimate.

3.76 L

$$\begin{array}{c}{C}_{1}{V}_{1}={C}_{2}{V}_{2}\\ \\ {V}_{1}=\phantom{\rule{0.2em}{0ex}}\frac{{C}_{2}{V}_{2}}{{C}_{1}}\phantom{\rule{0.2em}{0ex}}\end{array}$$

Since the concentration of the diluted solution 0.100 *M* is roughly one-sixteenth that of the stock solution (1.59 *M*), the volume of the stock solution is expected to be about one-sixteenth that of the diluted solution or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

$$\begin{array}{c}\\ {V}_{1}=\phantom{\rule{0.2em}{0ex}}\frac{\left(0.100\phantom{\rule{0.2em}{0ex}}M\right)\left(5.00\phantom{\rule{0.2em}{0ex}}\text{L}\right)}{1.59\phantom{\rule{0.2em}{0ex}}M}\phantom{\rule{0.2em}{0ex}}\\ {V}_{1}=0.314\phantom{\rule{0.2em}{0ex}}\text{L}\end{array}$$

Thus, 0.314 L of the 1.59-*M* solution is needed to prepare the desired solution. This result is consistent with the rough estimate.

0.261 L

By the end of this section, you will be able to:

- Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-per-million (ppm), and parts-per-billion (ppb)
- Perform computations relating a solution’s concentration and its components’ volumes and/or masses using these units

The previous section introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. This section will describe some other units of concentration that are commonly used in various applications, either for convenience or by convention.

Earlier in this chapter, percent composition was introduced as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures including solutions. The mass percentage of a solution component is defined as the ratio of the component’s mass to the solution’s mass, expressed as a percentage:

$$\text{mass percentage}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass of component}}{\text{mass of solution}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}100\%$$

Mass percentage is also referred to by similar names such as *percent mass, percent weight, weight/weight percent*, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).

Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 2.4) cites the concentration of its active ingredient, sodium hypochlorite (NaOCl), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl.

$$\%\phantom{\rule{0.2em}{0ex}}\text{glucose}=\phantom{\rule{0.2em}{0ex}}\frac{3.75\phantom{\rule{0.2em}{0ex}}\text{mg glucose}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{g}}{1000\phantom{\rule{0.2em}{0ex}}\text{mg}}\phantom{\rule{0.2em}{0ex}}}{5.0\phantom{\rule{0.2em}{0ex}}\text{g spinal fluid}}\phantom{\rule{0.2em}{0ex}}=0.075\%$$

The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%).

Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, the solute mass unit in the numerator was converted from mg to g to match the units in the denominator. Alternatively, the spinal fluid mass unit in the denominator could have been converted from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.

14.8%

For proper unit cancellation, the 0.500-L volume is converted into 500 mL, and the mass percentage is expressed as a ratio, 37.2 g HCl/g solution:

$$\text{500 mL solution}\phantom{\rule{0.2em}{0ex}}\left(\frac{1.19\phantom{\rule{0.2em}{0ex}}\text{g solution}}{\text{mL solution}}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{37.2\phantom{\rule{0.2em}{0ex}}\text{g HCl}}{100\phantom{\rule{0.2em}{0ex}}\text{g solution}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}221\phantom{\rule{0.2em}{0ex}}\text{g HCl}$$

This mass of HCl is consistent with our rough estimate of approximately 200 g.

282 mL

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)%:

$$\text{volume percentage}=\phantom{\rule{0.2em}{0ex}}\frac{\text{volume solute}}{\text{volume solution}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}100\%$$

$$\left(355\phantom{\rule{0.2em}{0ex}}\text{mL solution}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{70\phantom{\rule{0.2em}{0ex}}\text{mL isopropyl alcohol}}{100\phantom{\rule{0.2em}{0ex}}\text{mL solution}}\right)\phantom{\rule{0.2em}{0ex}}\left(\frac{0.785\phantom{\rule{0.2em}{0ex}}\text{g isopropyl alcohol}}{1\phantom{\rule{0.2em}{0ex}}\text{mL isopropyl alcohol}}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}195\phantom{\rule{0.2em}{0ex}}\text{g isopropyl alchol}$$

1.5 mol ethanol

“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 2.5).

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:

$$\begin{array}{l}\\ \text{ppm}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{ppm}\\ \text{ppb}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}\end{array}$$

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 2.6).

$$15\phantom{\rule{0.2em}{0ex}}\text{ppb}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{ppm}}{{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}=0.015\phantom{\rule{0.2em}{0ex}}\text{ppm}$$

The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. Since the volume of solution (300 mL) is given, its density must be used to derive the corresponding mass. 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:

$$\begin{array}{c}\\ \text{ppb}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass solute}}{\text{mass solution}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}\\ \\ \text{mass solute}=\phantom{\rule{0.2em}{0ex}}\frac{\text{ppb}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{mass solution}}{{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}\\ \\ \text{mass solute}=\phantom{\rule{0.2em}{0ex}}\frac{15\phantom{\rule{0.2em}{0ex}}\text{ppb}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}300\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1.00\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{mL}}\phantom{\rule{0.2em}{0ex}}}{{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{ppb}}\phantom{\rule{0.2em}{0ex}}=4.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{g}\end{array}$$

Finally, convert this mass to the requested unit of micrograms:

$$4.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{\mu g}}{{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{g}}\phantom{\rule{0.2em}{0ex}}=4.5\phantom{\rule{0.2em}{0ex}}\text{\mu g}$$

9.6 ppm, 9600 ppb

By the end of this section, you will be able to:

- Define and give examples of electrolytes
- Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes
- Relate electrolyte strength to solute-solvent attractive forces

When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.

Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure 2.7).

Water and other polar molecules are attracted to ions, as shown in Figure 2.8. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.

When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Even sparingly, soluble ionic compounds are strong electrolytes, since the small amount that does dissolve will dissociate completely.

Consider what happens at the microscopic level when solid KCl is added to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules surround individual K^{+} and Cl^{−} ions, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure 2.8 shows. Overcoming the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution as the ions transition from fixed positions in the undissolved compound to widely dispersed, solvated ions in solution.

Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton (H^{+} ion) to another molecule of water, yielding hydronium and hydroxide ions.

$${\text{H}}_{2}\text{O}\left(l\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons \phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}(aq)+{\text{OH}}^{\text{\u2212}}(aq)$$

In some cases, solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, an aqueous solution of HCl is a very good conductor, indicating that an appreciable concentration of ions exists within the solution.

Because HCl is an *acid*, its molecules react with water, transferring H^{+} ions to form hydronium ions (H_{3}O^{+}) and chloride ions (Cl^{−}):

This reaction is essentially 100% complete for HCl (i.e., it is a *strong acid* and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry.

Previous Citation(s)

Flowers, P., et al. (2019). Chemistry: Atoms First 2e. https://openstax.org/details/books/chemistry-atoms-first-2e (6.3-6.4, 11.2)

This content is provided to you freely by BYU Open Learning Network.

Access it online or download it at https://open.byu.edu/general_college_chemistry_2/chapter_2.