MCAT General Chemistry Review
Part I Review
Chapter 9: Solutions
What do first aid instant cold packs and sweet tea have in common? Not much, you might think, but both, in fact, demonstrate the same principles of solution chemistry. Instant cold packs contain two compartments, one holding water and the other ammonium nitrate. When you bend the pack, the two compartments break open, allowing the ammonium nitrate to dissolve into the water. Sweet tea is made by dissolving a large amount of sugar into strongly brewed tea—the sweetest of the sweet teas may approach 22 brix, which means that in every 100 grams of tea, there are 22 grams of sugar. (As a point of reference, this is about twice the sweetness of regular cola.) When the ammonium nitrate or the sugar dissolves in the water to produce the respective solutions, three events must take place. First, the intermolecular interactions between the molecules of ammonium nitrate (or between the sugar molecules) must be broken. Second, the intermolecular interactions between the molecules of water must be broken. Third, new intermolecular interactions between the ammonium nitrate and the water molecules (or between the sugar and the water) must be formed.
We know from our discussion in Chapter 3, Bonding and Chemical Interactions, that energy is required to break these intermolecular interactions and that energy is released when new ones are formed. For the creation of both the ammonium nitrate and sugar solutions, more energy is needed to break the original intermolecular interactions than is released when new intermolecular interactions are formed. Thus, the creation of both of these solutions is an endothermic process. Now, the formation of the ammonium nitrate solution is much more endothermic than the formation of the sugar solution. And this is why ammonium nitrate is useful in instant cold packs. When it dissolves in water, the system must absorb an amount of energy equal to 6.14 kcal/mol of ammonium nitrate. The heat is absorbed from the surrounding environment, and the pack feels cool to the touch.
Although the dissolution of sugar into water is not as strongly endothermic, we nevertheless have an intuitive understanding of the process’s endothermicity because we all know that the easiest way to dissolve lots of sugar into water (such as tea or coffee) is to heat up the water and then add the sugar. Because heating the water increases the solubility of sugar, it must be that the dissolution of sugar into water is an endothermic process (think application of Le Châtelier’s principle and changes in temperature from Chapter 5, Chemical Kinetics and Equilibrium).
In this chapter, our focus will be on the characteristics and behaviors of solutions. We have already begun our consideration of solutions in the previous chapter with our review of the colligative properties. We will now continue that review by examining the nature of solutions, the formation of aqueous solutions, measurements of solution concentration, and finally, the qualitative and quantitative evaluation of solution equilibria.
Nature of Solutions
Many important chemical reactions, both in the laboratory and in nature, take place in solution, including almost all reactions in living organisms. Solutions are homogenous (the same throughout) mixtures of two or more substances that combine to form a single phase, usually the liquid phase. The MCAT will focus generally on solids dissolved into liquid aqueous solutions, but it’s important to remember that solutions can be formed from different combinations of the three phases of matter. For example, gases can be dissolved in liquids (e.g., the carbonation of soda); liquids can be dissolved in other liquids (e.g., ethanol in water); solids can even be dissolved in other solids (e.g., metal alloys). Incidentally, gases “dissolved” into other gases can be thought of as solutions but are more properly defined as mixtures, because gas molecules don’t really interact all that much (one of the postulates of the kinetic molecular theory of gases). Just a point of minor clarification: All solutions are considered mixtures, but not all mixtures are considered solutions.
A solution consists of a solute (e.g., NaCl, NH3, C6H12O6, CO2, etc.) dissolved (dispersed) in a solvent (e.g., H2O, benzene, ethanol, etc.). The solvent is the component of the solution whose phase remains the same after mixing. If the two substances are already in the same phase (for example, a solution of two liquids), the solvent is the component present in greater quantity. And if the two same-phase components are in equal proportions in the solution, the component considered the solvent is the one that is more commonly identified as a solvent. Solute molecules move about freely in the solvent and interact with the solvent by way of interparticle forces such as ion–dipole, dipole–dipole, or hydrogen bonding. Dissolved solute molecules are also relatively free to interact with other dissolved molecules of different chemical identity; consequently, chemical reactions occur easily in solution.
Not to be confused with the path to eternal life, solvation is the electrostatic interaction between solute and solvent molecules. This is also known as dissolution, and when water is the solvent, it is called hydration, and the resulting solution is an aqueous solution. Solvation involves breaking intermolecular interactions between solute molecules and between solvent molecules and forming new intermolecular interactions between solute and solvent molecules.
When the new interactions (attractions) are stronger than the original ones, solvation is exothermic, and the process is favored at low temperatures. The dissolution of gases into liquids, such as CO2 into water, is an exothermic process because the only significant interactions that must be broken are those between water molecules. CO2, as a gas, demonstrates minimal intermolecular interaction, and thus the dissolution of CO2 gas into water is overall exothermic (and Le Châtelier’s principle tells us this is the reason that lowering the temperature of a liquid favors solubility of a gas in the liquid.) This is also why opening a two-liter container of warm soda is risky: The warm soda has very low solubility for the CO2 gas, and when the pressure is released upon twisting the cap open, a lot of the CO2 gas escapes (precipitates, comes out of the solution) with some amount of violence, carrying with it a lot of the liquid itself. We’ve all seen—or suffered—the aftermath of a warm soda too hastily opened.
When the new interactions (attractions) are weaker than the original ones, solvation is endothermic, and the process is favored at high temperatures. Most dissolutions are of this type. Two such examples have already been given: dissolving sugar or ammonium nitrate into water. Since the new interactions between the solute and solvent are weaker than the original interactions between the solute molecules and between the solvent molecules, energy (heat) must be supplied to facilitate the formation of these weaker, less stable interactions. Sometimes the overall strength of the new interactions is approximately equal to the overall strength of the original interactions. In this case, the overall enthalpy change for the dissolution is close to zero. These types of solutions approximate the formation of an ideal solution, for which the enthalpy of dissolution is equal to zero.
The spontaneity of dissolution is not dependent only upon the enthalpy change; solutions may form spontaneously for both endothermic and exothermic dissolutions. The second property that contributes to the spontaneity or nonspontaneity of a solution is the entropy change that occurs in the process. At constant temperature and pressure, entropy always increases upon dissolution. As with any process, the spontaneity or nonspontaneity of dissolution depends upon the change in Gibbs function: Spontaneous processes result in a decrease of free energy, while nonspontaneous processes result in an increase of free energy. Thus, whether dissolution will happen spontaneously or not depends upon both the change in enthalpy and change in entropy for the solute and solvent of the system.
Consider, for example, the formation of another common solution, one that we’ve mentioned a number of times in this book: sodium chloride, table salt, dissolved in water. When NaCl dissolves in water, its component ions dissociate from each other and become surrounded by water molecules. For this new interaction to occur, ion–ion interactions between the Na+ and Cl- must be broken, and hydrogen bonds between water molecules must also be broken. This step requires energy and is therefore endothermic. Because water is polar, it can interact with each of the component ions through ion–dipole interactions: The partially positive hydrogen end of the water molecules will surround the Cl- ions, and the partially negative oxygen end of the water molecules will surround the Na+ ions. The formation of these ion–dipole bonds is exothermic, but not as much so as the endothermicity of breaking the old ones (although it is quite close). As a result, the overall dissolution of table salt into water is endothermic (+0.93 J/mol) and favored at high temperatures.
We’ve considered the enthalpy change for the formation of a sodium chloride solution, and now we need to examine the entropy change. Remember that entropy can be thought of as the measure of the degree to which energy is disbursed throughout a system or the measure of the amount of energy distributed from the system to the surroundings at a given temperature. Another way to understand entropy is the measure of molecular disorder, or the number of energy microstates available to a system at a given temperature. When solid sodium chloride dissolves into water, the rigidly ordered arrangement of the sodium and chloride ions is broken up, as the ion–ion interactions are disrupted and new ion–dipole interactions with the water molecules are formed. The ions, freed from their lattice arrangement, have a greater number of energy microstates available to them (in simpler terms, they are freer to move around in different ways), and consequently, their energy is more distributed and their entropy increases. The water, however, becomes more restricted in its movement because it is now interacting with the ions. The number of energy microstates available to it (that is, the water molecules’ ability to move around in different ways) is reduced, so the entropy of the water decreases. But the increase in the entropy experienced by the dissolved sodium chloride is greater than the decrease in the entropy experienced by the water, so the overall entropy change is positive—energy is overall disbursed by the dissolution of sodium chloride in water. Because of the relatively low endothermicity and relatively large positive change in entropy, sodium chloride will spontaneously dissolve in liquid water.
We need to know more than just whether or not dissolution of a solute into a solvent will be spontaneous or nonspontaneous. We also want to know how much solute will dissolve into a given solvent. The solubility of a substance is the maximum amount of that substance that can be dissolved in a particular solvent at a particular temperature. When this maximum amount of solute has been added, the dissolved solute is in equilibrium with its undissolved state, and we say that the solution is saturated. If more solute is added, it will not dissolve. For example, at 18°C, a maximum of 83 g of glucose (C6H12O6) will dissolve in 100 mL of H2O. Thus, the solubility of glucose is 83 g/100 mL. If more glucose is added to an already saturated glucose solution, it will not dissolve but rather will remain in solid form, precipitating to the bottom of the container. A solution in which the proportion of solute to solvent is small is said to be dilute, and one in which the proportion is large is said to be concentrated. Note that both dilute and concentrated solutions are still considered unsaturated if the maximum equilibrium concentration (saturation) has not been reached.
The solubility of substances into different solvents is ultimately a function of thermodynamics. When the change in Gibbs function is negative at a given temperature for the dissolution of a given solute into a given solvent, the process will be spontaneous, and the solute is said to be soluble. When the change in Gibbs function is positive at a given temperature for the dissolution of a given solute into a given solvent, the process will be nonspontaneous, and the solute is said to be insoluble. Some solute/solvent systems have very large negative Gs, so dissolution is very spontaneous and a lot of solute can be dissolved into the solvent. Others have very small negative Gs, so dissolution is only slightly spontaneous and as a result only a little solute can be dissolved into the solvent. Those solutes that dissolve minimally in the solvent (usually water) are called the sparingly soluble salts.
The most common type of solution is the aqueous solution, in which the solvent is water. The aqueous state is denoted by the symbol (aq). Because aqueous solutions are so common and so important to biological systems (e.g., you), the MCAT focuses on them above all others. We wish we could tell you otherwise, but for Test Day, you are expected to remember the general solubility rules for aqueous solutions. We know, we know—they are not the easiest or most fun to remember, but no one ever promised you that this would always be fun. (We promised you that we would help you have some fun in this preparation process.) There are seven general solubility rules:
1. All salts of alkali metals are water soluble.
2. All salts of the ammonium ion (NH4+) are water soluble.
3. All chlorides, bromides, and iodides are water soluble, with the exceptions of Ag+, Pb2+, and Hg22+.
4. All salts of the sulfate ion (SO42-) are water soluble, with the exceptions of Ca2+, Sr2+, Ba2+, and Pb2+.
5. All metal oxides are insoluble, with the exception of the alkali metals and CaO, SrO, and BaO, all of which hydrolyze to form solutions of the corresponding metal hydroxides.
6. All hydroxides are insoluble, with the exception of the alkali metals and Ca 2+, Sr2+, and Ba2+.
7. All carbonates (CO32-), phosphates (PO43-), sulfides (S2-), and sulfites (SO32-) are insoluble, with the exception of the alkali metals and ammonium.
On the MCAT, there is one infallible solubility rule: All sodium salts are completely soluble, and all nitrate salts are completely soluble. Thus, if a problem gives you a concentration of sodium fluoride, you know that the compound is completely soluble in water. Sodium and nitrate ions are generally used as counterions to what is really chemically important; for example, if a pH problem gives you a sodium formate concentration of 0.10 M, it is really telling you that the concentration of the formate ion is 0.10 M, because the sodium ion concentration does not affect pH. The only time you need to worry about the nitrate ion concentration is in a redox reaction, for the nitrate ion can function—though only weakly—as an oxidizing agent; otherwise, merely focus on the cation as the chemically reacting species.
Because most solutions involve water as the solvent in the real world, it is not a surprise that they are common on the MCAT. These solubility rules are not bad to know, but memorizing them all may be a little excessive. It is never a bad thing to know facts, but being able to apply them is more important. Know rules 1 and 2 for sure and be aware of some of the more common insoluble exceptions, like Pb2+ and Ag+.
Ionic solutions are of particular interest to chemists because certain important types of chemical reactions—acid-base and oxidation-reduction reactions, for instance—take place in ionic solutions. It shouldn’t come as a surprise to you by now that if chemists take particular interest in this area of chemistry, so will the MCAT. In fact, acid-base and oxidation-reduction reactions are themselves important topics for the MCAT, so you can begin your review of those topics (Chapters 10 and 11) by reviewing with us now the characteristics and behaviors of ions.
CATIONS AND ANIONS
Ionic compounds are made up of positively charged cations and negatively charged anions. Ionic compounds are held together by the ionic bond, which is the force of electrostatic attraction between oppositely charged particles. The word cation (and especially the organic chemistry variant,carbocation) has tripped up unsuspecting students for years, who, having never heard the word spoken aloud before, almost invariably pronounce it as the last two syllables of “vacation” (as if a carbocation were some fun carbon-based holiday…we humbly admit to having made the mistake ourselves at a point in the less glorious distant past.)
The nomenclature of ionic compounds is based on the names of the component ions.
1. For elements (usually metals) that can form more than one positive ion, the charge is indicated by a Roman numeral in parentheses following the name of the element.
Fe2+ Iron (II)
Cu+ Copper (I)
Fe3+ Iron (III)
Cu2+ Copper (II)
2. An older but still commonly used method is to add the endings -ous or -ic to the root of the Latin name of the element to represent the ions with lesser or greater charge, respectively.
3. Monatomic anions are named by dropping the ending of the name of the element and adding -ide.
4. Many polyatomic anions contain oxygen and are therefore called oxyanions. When an element forms two oxyanions, the name of the one with less oxygen ends in -ite and the one with more oxygen ends in -ate.
5. When the series of oxyanions contains four oxyanions, prefixes are also used. Hypo- and per- are used to indicate less oxygen and more oxygen, respectively.
6. Polyatomic anions often gain one or more H+ ions to form anions of lower charge. The resulting ions are named by adding the word hydrogen or dihydrogen to the front of the anion’s name. An older method uses the prefix bi- to indicate the addition of a single hydrogen ion.
HCO3– Hydrogen carbonate or bicarbonate
HSO4– Hydrogen sulfate or bisulfate
H2PO4– Dihydrogen phosphate
Ionic species, by definition, have charge. Cations have positive charge, and anions have negative charge. Some elements are found naturally only in their charged forms, while others may exist naturally in the charged or uncharged state. Furthermore, some elements can have several different charges or oxidation states. Some of the charged atoms or molecules that you might commonly encounter on the MCAT include the active metals, the alkali metals (group IA) and the alkaline earth metals (group IIA), which have charge of +1 and +2, respectively, in the natural state. Many of the transition metals, such as copper, iron, and chromium, can exist in different positively charged states. Nonmetals, which are found on the right side of the periodic table, generally form anions. For example, all the halogens (Group VIIA) form monatomic anions with a charge of -1. All elements in a given group tend to form monatomic ions with the same charge (e.g., Group IA elements have charge of +1). Note that there are anionic species that contain metallic elements (e.g., MnO4- [permanganate] and CrO42- [chromate]); even so, the metals have positively charged oxidation states. (Also note that in the oxyanions of the halogens, such as ClO- and ClO2-, the halogen is assigned a positive oxidation state.) The trends of ionicity as we’ve described them here are helpful but are complicated by the fact that many elements have intermediate electronegativity and are consequently less likely to form ionic compounds and by the left-to-right transition from metallic to nonmetallic character.
Oxyanions of transition metals like the MnO4– and CrO42– ions shown here have an inordinately high oxidation number on the metal. As such, they tend to gain electrons in order to reduce this oxidation number and thus make good oxidizing agents. (See Chapter 11.)
In spite of the fact that ionic compounds are made of ions, solid ionic compounds tend to be poor conductors of electricity, because the charged particles are rigidly set in place by the lattice arrangement that serves as the basic framework for crystalline solids. In aqueous solutions, however, the lattice arrangement is disrupted by the ion–dipole interactions between the ionic constituents and the water molecules. The freed-up ions are now able to move around, and as a result, the solution of ions is able to conduct electricity. Solutes that enable their solution to carry currents are called electrolytes. The electrical conductivity of aqueous solutions is governed, then, by the presence and concentration of ions in the solution. Pure water, which has no ions other than the very few hydrogen ions and hydroxide ions that result from water’s low-level autodissociation, is a very poor conductor.
Because electrolytes ionize in solution, they will produce a larger effect on colligative properties (see Chapter 8) than one would expect from the given concentration.
The tendency of an ionic solute to dissociate into its constituent ions in water may be high or low. A solute is considered a strong electrolyte if it dissociates completely into its constituent ions. Examples of strong electrolytes include certain ionic compounds, such as NaCl and KI, and molecular compounds with highly polar covalent bonds that dissociate into ions when dissolved, such as HCl in water. A weak electrolyte, on the other hand, ionizes or hydrolyzes incompletely in aqueous solution, and only some of the solute is dissolved into its ion constituents. Examples include Hg2I2 (Ksp= 4.5 × 10-29), acetic acid and other weak acids, ammonia, and other weak bases (see Chapter 10, Acids and Bases). Many compounds do not ionize at all in aqueous solution, retaining their molecular structure in solution, which usually limits their solubility. These compounds are called nonelectrolytes and include many nonpolar gases and organic compounds, such as O2 (g), CO2 (g), and glucose.
Concentration denotes the amount of solute dissolved in a solvent. There are many different ways of expressing concentration, and different units have been standardized that you may encounter in everyday situations. For example, alcohol content in liquors like vodka, gin, or rum is expressed in volume percent (volume of solute divided by volume of solution times 100 percent). Alcoholic proof is twice the volume percent. The sugar content of orange juice and other fruit juices is measured in units of degrees Brix (°Bx), which is a weight (actually mass) percent: mass of glucose divided by mass of solution times 100 percent.
UNITS OF CONCENTRATION
On the MCAT test, you will work with concentrations of solutions commonly expressed as percent composition by mass, mole fraction, molarity, molality, and normality.
Percent Composition by Mass
The percent composition by mass (w/w%) of a solution is the mass of the solute divided by the mass of the solution (solute plus solvent), multiplied by 100 percent.
It is important to have a good idea of how to work with all of these ways of expressing concentration because more than one may show up on Test Day.
Example: What is the percent composition by mass of a salt water solution if 100 g of the solution contains 20 g of NaCl?
Solution: × 100 + 20% NaCl solution
The mole fraction (X) of a compound is equal to the number of moles of the compound divided by the total number of moles of all species with the system. The sum of the mole fractions in a system will always equal 1. Mole fraction is used to calculate the vapor pressure depression of a solution, as well as the partial pressures of gases in a system.
Example: If 92 g of glycerol is mixed with 90 g of water, what will be the mole fractions of the two components? (MW of H2O = 18; MW of C3H8O3 = 92.)
Note that for dilute solutions, the volume of the solution is approximately equal to the volume of solvent used, which simplifies our need for calculation on Test Day.
The molarity (M) of a solution is the number of moles of a solute per liter of solution. Solution concentrations are usually expressed in terms of molarity, and you will be working mostly with molarity on the MCAT. The molarity of a solution is written using brackets. Please note that the volume term in the denominator of molarity refers to the solution volume, not the solvent volume used to prepare the solution—although often the two values are close enough that we can approximate the solution volume by the solvent volume. We use molarity for the law of mass action, rate laws, osmotic pressure, pH and pOH, and the Nernst equation.
Example: If enough water is added to 11 g of CaCl2 to make 100 mL of solution, what is the molarity of the solution?
The molality (m) of a solution is the number of moles of solute per kilogram of solvent. For dilute aqueous solutions at 25°C, the molality is approximately equal to molarity, because the density of water at this temperature is 1 kilogram per liter (1 kg/L). However, note that this is an approximation and true only for dilute aqueous solutions. (As aqueous solutions become more concentrated with solute, their densities become significantly different from that of pure water; most water-soluble solutes have molecular weights significantly greater than that of water, so the density of the solution increases as the concentration increases.) You won’t use molality very often, so be mindful of the special situations when it is required: boiling point elevation and freezing point depression.
Example: If 10 g of NaOH are dissolved in 500 g of water, what is the molality of the solution?
We discussed the related concepts of gram equivalent weight, equivalents, and normality (N) in Chapter 4. We would urge you to review those relevant sections. The normality of a solution is equal to the number of equivalents of solute per liter of solution. An equivalent, or gram equivalent weight, is a measure of the reactive capacity of a molecule. Most simply, an equivalent is equal to a mole of charge.
To calculate the normality of a solution, you need to know for what purpose the solution is being used, because it is the concentration of the reactive species with which we are concerned. For example, in acid-base reactions, we are most concerned with the concentration of hydrogen ions; in oxidation-reduction reactions, we are most concerned with the concentration of electrons. Normality is unique among concentration units in that it is reaction-dependent. For example, in acidic solution, 1 mole of the permanganate ion (MnO4-) will readily accept 5 moles of electrons, so a 1 M solution would be 5 N. However, in alkaline solution, 1 mole of permanganate will accept only 3 moles of electrons, so in alkaline solution, a 1 M permanganate solution would be 3 N.
Simple ideas on Test Day will make things easier. So, when you come across normality you can think of it as “molarity of the stuff of interest” in the reaction.
A solution is diluted when solvent is added to a solution of high concentration to produce a solution of lower concentration. The concentration of a solution after dilution can be conveniently determined using the equation:
MiVi = MfVf
where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values, respectively.
This equation is worthy of memorization. Note that it works for any units of concentration, not just molarity, if we replace the M with C for concentration.
Example: How many mL of a 5.5 M NaOH solution must be used to prepare 300 mL of a 1.2 M NaOH solution?
Our last topic for this chapter on solutions picks up on a theme that we began discussing in Chapter 5, Chemical Kinetics and Equilibrium. The process of solvation, like other reversible chemical and physical processes, tends toward an equilibrium position defined as the lowest energy state of a system under given conditions of temperature and pressure. Related to determinations of equilibrium is the characterization of processes as spontaneous or nonspontaneous: Systems tend to move spontaneously toward the equilibrium position, but any movement away from equilibrium is nonspontaneous. In the process of creating a solution, the equilibrium is defined as the saturation point and the solute concentration is at its maximum value for the given temperature and pressure. Immediately after solute has been introduced into a solvent, most of the change taking place is dissociation, because no dissolved solute is initially present. However, once solute is dissolved, the reverse process, precipitation of the solute, will begin to occur. When the solution is dilute (unsaturated), the thermodynamically favored process is dissolution, and initially, the rate of dissolution will be greater than the rate of precipitation. As the solution becomes more concentrated and approaches saturation, the rate of dissolution lessens while the rate of precipitation increases. Eventually, the saturation point of the solution is reached, and the solution exists in a state of dynamic equilibrium for which the rates of dissolution and precipitation are equal and the concentration of dissolved solute reaches a steady state (that is, constant) value. Neither dissolution nor precipitation is more thermodynamically favored at equilibrium than the other (because favoring either one of them would necessarily result in the solution no longer being in a state of equilibrium), so the change in free energy is zero, as is the case for all systems at equilibrium.
An ionic solid introduced into a polar solvent dissociates into its component ions, and the dissociation of such a solute in solution may be represented by
AmBn (s) mAn+ (aq) + nBm- (aq)
On Test Day, when you are working through any problem of solution chemistry, the first step you must take is to write out the balanced dissolution equation for the ionic compound in question. This first step is essential for correctly calculating solubility product constant, ion product, molar solubility, or common ion effect. In other words, it is the essential first step for nearly every solution chemistry problem you will see on the MCAT.
THE SOLUBILITY PRODUCT CONSTANT
Most solubility problems on the MCAT deal with solutions of sparingly soluble salts, which are ionic compounds that have very low solubility in aqueous solutions. You may wonder why any ionic compound would not be highly soluble in water. The determination for the degree of solubility is the relative changes in enthalpy and entropy associated with the dissolution of the ionic solute at a given temperature and pressure. One common sparingly soluble salt is silver chloride, AgCl, which dissolves according to the following equation:
AgCl (s) Ag+ (aq) + Cl- (aq)
The law of mass action can be applied to a solution at equilibrium; that is to say, when the solution is saturated and the solute concentration is maximum and dynamically stable. For a saturated solution of the ionic compound with the formula AmBn, the equilibrium constant for its solubility in aqueous solution, called the solubility product constant Ksp, can be expressed by
Ksp = [An+]m[Bm-]n
where the concentrations of the ionic constituents are equilibrium (saturation) concentrations. For example, we can express the Ksp of silver chloride as follows:
Ksp = [Ag+] [Cl-]
You’ll notice that for the law of mass action of solutions, the denominator seems to be missing. Well, if you think back to our discussion of the properties of the equilibrium constant in Chapter 5, you’ll remember that we don’t include the concentration of the pure solids or pure liquids. Since the silver chloride solution was formed by adding pure solid silver chloride to pure water, neither the solid silver chloride nor the water is included.
Solubility product constants, like all other equilibrium constants (Keq, Ka, and Kb) are temperature-dependent. When the solution consists of a gas dissolved into a liquid, the value of the equilibrium constant, and hence the “position” of equilibrium (saturation), will also depend on pressure. Generally speaking, the solubility product constant increases with increasing temperature for nongas solutes and decreases for gas solutes. Higher pressures favor dissolution of gas solutes and therefore the Ksp will be larger for gases at higher pressures than at lower ones. This last point is especially relevant to deep-sea divers. Because gases become more soluble in solution as pressure increases, a diver who has spent time at significant depths under water will have more nitrogen gas dissolved in her blood. (Nitrogen gas is the main inert gas in the air we breathe.) If she rises to the surface too quickly, the abrupt decompression will lead to an abrupt decrease in gas solubility in the plasma, resulting in the formation of nitrogen gas bubbles in her bloodstream. The gas bubbles can get lodged in the small vasculature of the peripheral tissue, mostly around the large joints of the body, causing pain and tissue damage (hence the name of the condition is the “bends”). The condition is painful and dangerous, and can be fatal if not properly prevented or treated.
On the MCAT, if you remember that Ksp is just a specialized form of Keq, then you can simplify a lot of problems by using the same concepts that you do for all equilibria, including Le Châtelier’s principle.
As solute dissolves into solvent the system approaches saturation, at which point no more solute can be dissolved and any excess will precipitate to the bottom of the container. You may not know whether the solution has reached saturation, and so to determine “where” the system is with respect to the equilibrium position, you will calculate a value called the ion product (I.P.), which is analogous to the reaction quotient Q for chemical reactions. The ion product equation has the same form as the equation for the solubility product constant. The difference is that the concentrations that you use are the concentrations of the ionic constituents at that given moment in time.
I.P. = [An+]m[Bm-]n
If the solution is supersaturated, Qsp > Ksp, precipitation will occur. If the solution is unsaturated, Qsp < Ksp, the solute will continue to dissolve. If the solution is saturated, Qsp = Ksp, then the solution is at equilibrium.
where the concentrations are not necessarily equilibrium (saturation) concentrations. As with the reaction quotient Q, the utility of the I.P. lies in comparing its value to that attained at equilibrium, in this case, the known Ksp. Each salt has its own distinct Ksp at a given temperature and pressure. If, at a given set of temperature and pressure conditions, a salt’s I.P. is less than the salt’s Ksp, then the solution is not yet at equilibrium and we say that it is unsaturated. For unsaturated solutions, dissolution is thermodynamically favored over precipitation. If the I.P is greater than the Ksp, then the solution is beyond equilibrium and we say that it is supersaturated. It’s possible to create a supersaturated solution by dissolving solute into a hot solvent and then slowly cooling the solution. A supersaturated solution is thermodynamically unstable, and any disturbance to the solution, like the addition of more solid solute or other solid particles or further cooling of the solution, will cause spontaneous precipitation of the excess dissolved solute. If the calculated I.P. is equal to the known Ksp, then the solution is at equilibrium, the rates of dissolution and precipitation are equal, and the concentration of solute is at the maximum (saturation) value.
Example: The molar solubility of Fe(OH)3 in an aqueous solution was determined to be 4.5 × 10-10 mol/L. What is the value of the Ksp for Fe(OH)3?
Every slightly soluble salt of general formula MX3 will have Ksp = 27x4, where x is the molar solubility.
Solution: The molar solubility (the solubility of the compound in mol/L) is given as 4.5 × 10-10 M. The equilibrium concentration of each ion can be determined from the molar solubility and the balanced dissociation reaction of Fe(OH)3. The dissociation reaction is:
Example: What are the concentrations of each of the ions in a saturated solution of PbBr2, given that the Ksp of PbBr2 is 2.1 × 10-6? If 5 g of PbBr2 are dissolved in water to make 1 L of solution at 25°C, would the solution be saturated, unsaturated, or supersaturated?
Every slightly soluble salt of general formula MX2 will have Ksp = 4x3, where x is the molar solubility.
Solution: The first step is to write out the dissociation reaction:
PbBr2(s) Pb2+(aq) + 2Br-(aq)
Ksp = [Pb2+][Br-]2
Let x equal the concentration of Pb2+. Then 2 x equals the concentration of Br– in the saturated solution at equilibrium (as [Br-] is 2 times [Pb2+]).
(x)(2x)2 = 4x3
2.1 × 10-6 = 4x3
Every slightly soluble salt of general formula MX will have Ksp = x2, where x is the molar solubility.Ksp =x2, where x is
Solving for x, the concentration of Pb2+ in a saturated solution is 8.07 × 10-3 M and the concentration of Br– (2x) is 1.61 × 10-2 M. Next, we convert 5 g of PbBr2 into moles:
1.36 × 10-2 mol of PbBr2 is dissolved in 1 L of solution, so the concentration of the solution 1.36 × 10-2 M. Because this is higher than the concentration of a saturated solution, this solution would be supersaturated.
THE COMMON ION EFFECT
The solubility of a substance varies depending on the temperature of the solution, the solvent, and, in the case of a gas-phase solute, the pressure. Solubility is also affected by the addition of other substances to the solution.
One of the more common solution chemistry problems on the MCAT is calculation of the concentration of a salt in a solution that already contains a common ionic constituent. The solubility of a salt is considerably reduced when it is dissolved in a solution that already contains one of its constituent ions compared to its solubility in the pure solvent. This reduction in molar solubility is called the common ion effect. Molar solubility (M ) is the concentration, in moles per liter (mol/L), of the solute in the solution at equilibrium at a given temperature. If X moles of AmBn (s) can be dissolved in Y liters of solution to reach saturation, then the molar solubility of AmBn (s) is X mol/Y L. Let us repeat the important effect of the common ion: Its presence results in a reduction in the molar solubility of the salt. Note well that the presence of the common ion has no effect on the value of the solubility product constant for the salt. For example, if a salt such as CaF2 is dissolved into a solvent already containing Ca2+ ions (from some other salt, perhaps CaCl2), the solution will dissolve less CaF2 compared to the amount that would be dissolved in the pure solvent before the I.P. equals Ksp. The common ion effect is really nothing other than Le Châtelier’s principle in action. Because the solution already contains one of the constituent ions from the right side of the dissociation equilibrium, we can see that the system will shift away from that side toward the left side, where we find the solid salt. A solution system shifting toward the left (solid salt reactant) is not going to favor dissolution. As a result, molar solubility for the solid is reduced, and less of the solid dissolves in the solution (for the same Ksp).
Example: The Ksp of Agl in aqueous solution is 1 × 10-16 mol/L. If a 1 × 10-5 M solution of AgNO3 is saturated with AgI, what will be the final concentration of the iodide ion?
Solution: The concentration of Ag+ in the original AgNO3 solution will be 1 × 10-5 mol/L. After AgI is added to saturation, the iodide concentration can be found by this formula:
If the AgI had been dissolved in pure water, the concentration of both Ag+ and I- would have been 1 × 10-8 mol/L. The presence of the common ion, silver, at a concentration 1,000 times higher than what it would normally be in a silver iodide solution has reduced the iodide concentration to 1,000 of what it would have been otherwise. An additional 1 × 10-11 mol/L of silver will, of course, dissolve in solution along with the iodide ion, but this will not significantly affect the final silver concentration, which is much higher.
Our review of solution chemistry has provided an opportunity for us to consider the nature of solutions, solutes, and solvents and the manner of interaction between solutes and solvents in the formation of solutions. We reviewed solubility and the rules that reflect the solubility of common compounds in water. The different ways of expressing the amount of solute in solution were identified, and examples were given for each unit of concentration, including percent composition, mole fraction, molarity, molality, and normality. Finally, we reviewed the thermodynamic principles of solution equilibria and defined unsaturated, saturated, and supersaturated solutions with reference to ion product (I.P.) and solubility product constant (Ksp), as well as the common ion effect from the perspective of Le Châtelier’s principle for a solution at equilibrium.
CONCEPTS TO REMEMBER
Solutions are homogenous mixtures of two or more substances that combine to form a single phase, generally the liquid phase. The most important kind of solution for the MCAT is the aqueous solution. Solvents dissolve solutes by a process of surrounding the solute particles and interacting with them by way of electrostatic forces; this is called solvation.
Most dissolutions are endothermic. However, the dissolution of gas into liquid is exothermic.
Solubility is the maximum amount of substance that can be dissolved in a particular solvent at a particular temperature. Molar solubility is the molar concentration of solute in a saturated solution.
Units of solution concentration include percent composition by mass (mass of solute divided by mass of solution times 100%), mole fraction (moles of solute divided by total number of moles of substances in solution), molarity (moles of solute divided by liters of solution), molality (moles of solute divided by kilograms of solvent), and normality (number of equivalents divided by liters of solution).
A saturated solution is in equilibrium for that particular temperature. Ksp is the solubility product constant for a given solute in a given solvent at a given temperature.
Calculation of the ion product (I.P.), followed by comparison to the known Ksp, helps to determine if a solution is unsaturated (I.P < Ksp), saturated (I.P = Ksp), or supersaturated (I.P. > Ksp).
When an ionic compound is dissolved into a solution that already contains one of the constituent ions, the molar solubility for that ionic compound will be significantly decreased from the value normally demonstrated by the same ionic compound in the pure solvent at the same temperature. This is the common ion effect. It is an application of Le Châtelier’s principle to solutions.
EQUATIONS TO REMEMBER
MiVi = MfVf (the dilution law)
Ksp = [An+]m[Bm-]n for AmBn (s) mAn+ (aq) + nBm- (aq)
I.P. = [An+]m[Bm-]n
1. An aqueous solution was prepared by mixing 70 grams of sugar (C12H12O11) into 100 grams of water. The solution has a boiling point of 101.11°C. What is the molar mass of the solute? (Kb = 0.512°C.)
A. 322.58 g/mol
B. 32.26 g/mol
C. 123.24 g/mol
D. 233.59 g/mol
2. Which phase of solvent and solute, respectively, can form a solution?
I. Solid solvent, gaseous solute
II. Solid solvent, solid solute
III. Gaseous solvent, gaseous solute
A. I and II only
B. II and III only
C. I and III only
D. I, II, and III
3. Two organic liquids, pictured in the figure below, are combined to form a solution. Based on the structures, will the solution closely obey Raoult’s law?
A. Yes, the liquids differ due to the additional methyl group on toluene and, therefore, will not deviate from Raoult’s law.
B. Yes, the liquids are very similar and, therefore, will not deviate from Raoult’s law.
C. No, the liquids differ due to the additional methyl group on toluene and, therefore, will deviate from Raoult’s law.
D. No, the liquids both contain benzene rings, which will interact with each other and cause deviation from Raoult’s law.
4. The diagram in Figure 1 shows two arms separated by an impermeable membrane. What would happen to the level of liquid in the two branches if the membrane were replaced with a semipermeable membrane that allowed water molecules to move across the membrane?
A. The level on the right would decrease, and the level on the left would increase.
B. The level would remain the same on both sides.
C. The level on the right would increase, and the level on the left would decrease.
D. The level on the right would stay the same, and the level on the left would increase.
5. The process of formation of a liquid solution can be better understood by breaking the process into three steps:
1. Breaking the solution into individual components
2. Making room for the solute in the solvent by overcoming intermolecular forces in the solvent
3. Allowing solute–solvent interactions to occur to form the solution
The overall energy change to form a solution can be estimated by taking the sum of each of the three steps. For steps 1 through 3, will each step most likely be endothermic or exothermic? The order for each answer choice is step 1, step 2, followed by step 3.
A. Endothermic, exothermic, endothermic
B. Exothermic, endothermic, endothermic
C. Exothermic, exothermic, endothermic
D. Endothermic, endothermic, exothermic
6. The entropy change when a solution forms can be expressed by the term S°soln. When an ion dissolves and water molecules are ordered around it, the ordering would be expected to make a negative contribution to S°soln. An ion that has more charge density will have a greater hydration effect, or ordering of water molecules. Based on this information, which of the following compounds will have the most negative S°soln?
7. A 0.01 M solution of a nonelectrolyte has an osmotic pressure of 15.0 mm Hg. What is the osmotic pressure of a 0.02 M solution of Mg(NO3)2? The temperature of both solutions is the same.
A. 7.5 mm Hg
B. 30 mm Hg
C. 45 mm Hg
D. 90 mm Hg
8. One hundred grams of sugar are dissolved in a cup of hot water at 80°C. The cup of water contains 300.00 mL of water. What is the mass percentage of sugar in the resulting solution? (Sugar = C12H22O11, density of water at 80°C = 0.975 g/mL.)
9. Which of the following combinations of liquids would be expected to have a vapor pressure higher than the vapor pressure that would be predicted by Raoult’s law?
A. Ethanol and hexane
B. Acetone and water
C. Isopropanol and methanol
D. Nitric acid and water
10. The salt KCl is dissolved in a beaker of water that you are holding. You can feel the solution cool as the KCl dissolves. From this observation you conclude that
A. S°soln is large enough to overcome the unfavorable H°soln.
B. KCl is mostly insoluble in water.
C. S°soln must be negative when KCl dissolves.
D. Boiling point elevation will occur in this solution.
11. Which of the following will give the greatest increase in the boiling point of water when it is dissolved in 1.00 kg H2O?
A. 0.46 mol calcium sulfate
B. 0.54 mol iron (III) nitrate
C. 1.09 mol acetic acid
D. 1.11 mol sucrose
12. At sea level and 25°C, the solubility of oxygen gas in water is 1.25 × 10-3 M. In a city in the United States that lies high above sea level, the atmospheric pressure is 0.800 atm. What is the solubility of oxygen in water in this city?
A. 1.05 × 10-3 M
B. 1.56 × 10-3 M
C. 1.00 × 10-3 M
D. 1.25 × 10-3 M
13. Lead is a dangerous element that exists in the environment in large quantities due to man-made pollution. Lead poisoning has many symptoms, including mental retardation in children. If a body of water is polluted with lead ions at 30 ppb (parts per billion), what is the concentration of lead in molarity? (Density of water is 1 g/mL, ppb equals grams per 109 grams of solution.)
A. 6.2 × 10-7 M Pb2+
B. 1.4 × 10-10 M Pb2+
C. 1.4 × 10-7 M Pb2+
D. 6.2 × 10-6 M Pb2+
14. Which of the following is/are correct?
I. NaF is an electrolyte.
II. Glucose is a nonelectrolyte.
III. CH3OH is a weak electrolyte.
IV. CH3CH2COOH is a weak electrolyte.
A. I, III, and IV only
B. I and II only
C. II, and IV only
D. I, II, and IV only
15. Which one of the following is not a colligative property?
A. Boiling point elevation
B. Vapor pressure of a mixture
C. Osmotic pressure
D. Entropy of dissolution
16. The following equilibrium exists when AgBr is in solution:
Calculate the solubility of AgBr in g/L in a solution of 0.0010 M NaBr.
A. 7.7 × 10-13 g/L
B. 8.3 × 10-12 g/L
C. 7.7 × 10-10 g/L
D. 8.3 × 10-8 g/L
17. When ammonia, NH3, is a solvent, complex ions can form. For example, dissolving AgCl in NH3 will result in the complex ion Ag(NH3)2+. What effect would you expect the formation of complex ions to have on the solubility of a compound like AgCl in NH3?
A. The solubility of AgCl will increase, because complex ion formation will cause more ions to exist in solution, which interact with AgCl to cause it to dissociate.
B. The solubility of AgCl will increase, because complex ion formation will consume Ag+ molecules and cause the equilibrium to shift away from solid AgCl.
C. The solubility of AgCl will decrease, because Ag+ ions are in complexes and the Ag+ ions that are not complexed will want to associate with Cl- to form solid AgCl.
D. The solubility of AgCl will decrease, because complex ion formation will consume Ag+ molecules and cause the equilibrium to shift toward the solid AgCl.
18. Detergents are compounds that are dissolved in water. However, they are also able to dissolve hydrophobic stains, such as oil and grease in clothing and other fabrics. How are these compounds able to fulfill both hydrophilic and hydrophobic functions?
A. They contain a hydrophobic core molecule encased in a hydrophilic shell.
B. They can ionize into two parts; one part is ionic, and the other part is hydrophobic.
C. They have two states; in water they are ionic, and in hydrophobic solvents they form nonpolar ring structures.
D. They have two functionally distinct parts; one side is a hydrophobic chain, and the other end is polar and ionic.
Small Group Questions
1. How does the presence of common ions in solution affect Ksp? I.P. (Qsp)?
2. Discuss the circumstances that permit solvation. What forces are involved?
Explanations to Practice Questions
The equation Tb = iKbm can be used to solve this problem. The change in boiling point is found by subtracting the boiling point of water (the solvent), 100°C, from the elevated boiling point, 101.11°C. Using the given value for Kb, we solve for the molality of the solution and get 2.17 moles/kg. Convert to grams by dividing by 1,000 and then multiply by the mass of the solution, 100 g, to get the moles of solute. To obtain the molar mass, divide the mass of the solute, 70 g, by the number of moles, 0.217 moles, to get a molar mass of 322.58 g/mole. (B) would indicate a problem in unit conversions because the answer is off by an order of magnitude. (C) and (D) are wrong and would result from multiplying 1.11 by the Kb instead of dividing. (Note: Sugar does not dissociate in water, thus the van’t Hoff factor (i) is equal to one.)
All three choices can make a solution as long as the two components create a mixture that is of uniform appearance (homogenous). Hydrogen in platinum is an example of a gas in a solid. The air we breathe is an example of a homogenous mixture of a gas in a gas. Brass and steel are examples of homogenous mixtures of solids.
Benzene and toluene are both organic liquids and have very similar properties. They are both nonpolar and are almost exactly the same size. Raoult’s law states that ideal solution behavior is observed when solute–solute, solvent–solvent, and solute–solvent interactions are very similar. Therefore, benzene and toluene in solution will be predicted to behave as a nearly ideal solution. (A) states that the liquids would follow Raoult’s law because they are different. It is true that the compounds are slightly different, but the difference is negligible in terms of Raoult’s law. (C) and (D) are incorrect because they state that the solution would not obey Raoult’s law.
If the membrane became permeable to water, then water molecules would move to the side with the higher solute concentration, according to the principles of osmosis. The left side has a higher concentration of solute (NaCl), so water will move toward the left in an attempt to balance the concentrations of each side. The level on the left will rise because of the excess water molecules, and the level on the right will fall due to a loss of water molecules.
The first step will most likely be endothermic, because energy is required to break molecules apart. The second step is also endothermic, because the intermolecular forces in the solvent must be overcome to allow incorporation of solute particles. The third step will most likely be exothermic, because polar water molecules will interact with the dissolved ions and release energy.
CaS will cause the most negative S°soln because the Ca2+ and S2- ions have the highest charge density compared to the other ions. All of the other ions have charges of +1 or -1, whereas Ca2+ and S2- each have charges with an absolute value of 2. To arrange all four species in order of highest to lowest charge density, we’d have to take ion size into account. Smaller ions have higher charge densities. For example, LiF will have a higher charge density than KCl. It follows that the S°soln is more negative for LiF than for KCl.
A nonelectrolyte solution will not dissociate into ions in solution. Its effective molarity in solution will be the same as the number of moles that were dissolved. On the other hand, an electrolyte like Mg(NO3)2 will dissociate into three ions (Mg and 2NO3-). The effective molarity, which is important for colligative properties, will be three times the number of moles that were dissolved. Osmotic pressure is a colligative property and will, therefore, be three times larger for Mg(NO3)2 compared to a nonelectrolyte. The molarity of Mg(NO3)2, 0.02 M, is two times larger than the nonelectrolyte solution (0.01 M). The nonelectrolyte’s osmotic pressure will have to be multiplied by three and then by two for Mg(NO3)2: 15 mm Hg × 3 × 2 equals 90 mm Hg. We can use the osmotic pressure formula to check our work.
Osmotic pressure: = MRT
R terms cancel, and we’re solving for 2, so we’re left with
The mass percent of a solute equals the mass of the solute divided by the mass of the total solution. To find the mass of the solution, first find the mass of the solvent, water. Multiplying the volume of the solution by the density gives a mass of 292.5 grams of water. Adding 100 grams of sugar yields a solution with a mass of 392.5 grams. Next, divide 100 grams of sugar by 392.5 grams and multiply by 100 to get a percentage. (A) can be arrived at if water’s density at 80°C is assumed to be 1 g/mL. If we had forgotten to add the solute’s mass to the solvent’s, we’d have calculated 34.2 (100/292.5) percent, which is (D). (C) neglects both the addition step and the correct density value (100/300 = 33.3%).
Mixtures that have a higher vapor pressure than predicted by Raoult’s law have stronger solvent–solvent and solute–solute interactions than solvent–solute interactions. Therefore, particles do not want to stay in solution and more readily evaporate, creating a higher vapor pressure than an ideal solution. Two liquids that have different properties, like hexane (hydrophobic) and ethanol (hydrophilic, small), would not have many interactions with each other to cause positive deviation. (B) and (C) are composed of liquids that are similar to one another and that would not show significant deviation from Raoult’s law because they neither attract nor repel one another. (D) contains two liquids that would interact well with each other, causing a negative deviation from Raoult’s law. When attracted to one other, liquids prefer to stay in liquid form and have a lower vapor pressure than predicted by Raoult’s law.
Dissolution is governed by enthalpy and entropy, which are related by the equation G°soln = H°soln - °soln. The cooling of the solution indicates that heat is used up in this bond-breaking reaction. In other words, dissolution is endothermic, and H is positive. The reaction is occurring spontaneously, so G must be negative. The only way that a positive H can result in a negative G is if entropy, S, is a large, positive value. If has a larger absolute value than H, G will be negative. Conceptually, that means that the only way the solid can dissolve is if the increase in entropy is great enough to overcome the decrease in enthalpy. (B) is incorrect because it is clearly stated in the question stem that KCl dissolves. (C) is incorrect because S°soln must be positive in order for KCl to dissolve. Finally, (D) is incorrect because solute dissolution would cause boiling point to elevate, not depress. Additionally, it is not a piece of evidence that could be found simply by observing the beaker’s temperature change.
The equation to determine the change in boiling point of a solution is as follows: Tb = mb(Kb). mb is the molality of the solution, and Kb is the boiling point elevation constant. In this case, the solvent is always water, so Kb will be the same for each solution and we don’t need it to find our answer. What we do need to know is how many particles dissociate from each original species. This is referred to as the van’t Hoff factor (i) and is multiplied by our molality to demonstrate an effective molality. We’ll use effective molality values to determine which will cause the greatest change in boiling point.
It looks like we need to decide between iron (III) nitrate and acetic acid. No more calculations are necessary because the fact that acetic acid is a weak acid tells us that only a few particles will dissociate into H+ and acetate-. So the x that we are curious about is most likely closer to 1.09 than 2.18. We can certainly conclude that it will be less than 2.16. Iron (III) nitrate will have the largest effect on boiling point.
The solubility of gases in liquids is directly proportional to the atmospheric pressure. Therefore, we should expect a decrease in solubility upon experiencing the decreased atmospheric pressure in Denver. Because 0.800 atm is 80% of the pressure at sea level (1 atm), oxygen’s solubility will be 80% of 1.25 × 10-3, which yields 1.00 × 10-3 M. (A) is a miscalculation, (B) suggests that pressure and solubility are inversely related, and (D) implies that atmospheric temperature will not affect solubility.
30 ppb of Pb2+ is equivalent to 30 grams of Pb2+ in 109 grams of solution. Water’s density will help us convert from mass to volume. Dividing by the molar mass of Pb2+, 207 g/mole, will result in Pb2+’s molarity, 1.4 × 10-7 M.
An electrolyte is a molecule that dissociates into free ions and behaves as an electrically conductive medium. Number I, NaF, is an electrolyte because it dissociates to form the ions Na+ and F-. Number II, glucose, is a nonelectrolyte because it has a ring structure that dissolves but does not dissociate. Number III, CH3OH, has a pKa of approximately 15 and is not likely to ionize in solution. Therefore, it is not an electrolyte. Number IV, acetic acid, is a weak acid (pKa = 4.7) and also a weak electrolyte because it will partially ionize in solution.
A colligative property depends solely upon the number of molecules and disregards the identity of the molecules. (A), (B), and (C) are all properties based on the composition of a solution, determined by the number of molecules that are dissolved in the solution. (D), the entropy of dissolution, will depend on the chemical properties of the substance, such as charge density and electron affinity. Therefore, the entropy of dissolution is not a colligative property.
The solubility of AgBr can be solved by using the Ksp value given in the equation.
AgBr Ag+ + Br-
Ksp = [Ag+][Br-] = 7.7 × 10-13
Assuming that the same amount of cations and anions dissolve, we can refer to Ag+ and Br- both as x.
Ksp = [x][x] = 7.7 × 10-13
However, there is already 0.001 M Br- in solution. To account for that, we’ll add it in to the Ksp equation.
Ksp = [x][x + 0.001 M] = 7.7 × 10-13
We know 0.001 is much bigger than the square root of 10-13. In other words, it’s much bigger than x, and x is insignificant in comparison. So we’ll approximate Br-’s concentration as 0.001 M and solve for silver. The solubility of Ag+ is the same as AgBr’s because one molecule of Ag+ dissociates per molecule of AgBr.
We find that [Ag+] is 7.7 × 10-10 M Ag+. This value can be converted to g/L by multiplying by the molar mass of silver, 107.9 grams/mole.
(7.7 × 10-10 mol/L)(107.9 g/mol) = 8.3 × 10-8 g/L
Formation of complex ions between silver ions and ammonia will cause more molecules of solid AgCl to dissociate. The equilibrium is driven toward dissociation, because the Ag+ ions are essentially being removed from solution when they complex with ammonia. This rationale is based upon Le Châtelier’s principle, stating that when a chemical equilibrium experiences a change in concentration, the system will shift to counteract that change. (A) is incorrect because the complex ions may interact with AgCl but this is not the major reason for the increased solubility. (C) and (D) are incorrect because the solubility of AgCl will increase, not decrease.
Detergents are compounds that contain a long hydrophobic chain with a polar functional group on one end. The long hydrophobic chains can surround grease and oil droplets, while the polar heads face outward and carry the particles in a solution of water. (A) describes micelles, which are very different in configuration. (B) is internally inconsistent because ionization creates two ionic particles. (C) is incorrect because although multiple detergent molecules form a spherelike shape with oil or grease droplets enclosed, the individual molecules themselves do not form ring structures.