1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
The numbered diagrams are possible representations of one cation and one anion of the soluble ionic solid MX, with nearby water molecules, after MX has dissolved in water. In each water molecule, the larger circle represents the O atom. Which diagram best represents the interaction of the dissolved ions with the solvent?
Answer and reasoning
ADiagram 1Correct Dissolved ions are separated from one another, and each is surrounded by water molecules held by ion-dipole attractions: the partially negative O atoms of water are nearest the cation M⁺, and the partially positive H atoms are nearest the anion X⁻. This interaction of the dissolved species with the solvent is one of the factors in ΔG° of dissolution.
BDiagram 2 A student who has the partial charges of water the wrong way round picks the box in which H atoms are nearest M⁺ and O atoms are nearest X⁻. The O atom of water is partially negative and is attracted to the cation; the H atoms are partially positive and are attracted to the anion.
CDiagram 3 A student who treats water as a passive medium picks the box in which the ions are apart but the water molecules have no particular orientation. Water molecules are polar and are attracted to the ions, so those nearest an ion turn the oppositely charged end toward it.
DDiagram 4 A student who thinks a dissolved ionic compound remains as intact MX units picks the box in which M⁺ and X⁻ stay in contact. A soluble ionic solid dissolves as separate ions, each with its own surrounding water molecules.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.6.A.1 Free energy of dissolution, ΔG° Fix
Free energy of dissolution, ΔG°
The standard free energy change when a substance dissolves. It reflects three factors: the breaking of the interactions that hold the solid together, the reorganization of the solvent around the dissolved species, and the interaction of the dissolved species with the solvent. Each factor has an enthalpic and an entropic contribution.
Breaking the interactions that hold the solid together
Separating the particles of the solid requires energy (a positive contribution to ΔH°) and lets the particles become more dispersed (a positive contribution to ΔS°).
Reorganization of the solvent
Solvent molecules must move apart to make room for the dissolved species, which requires energy, and they become arranged around the dissolved species. Water molecules held in oriented positions around an ion are less free to move, which is a negative contribution to ΔS°; the effect is largest for small, highly charged ions.
Interaction of the dissolved species with the solvent
Attractions form between the dissolved species and the solvent molecules, for example ion-dipole attractions between ions and water, with the partially negative O atoms of water toward cations and the partially positive H atoms toward anions. Forming these attractions releases energy (a negative contribution to ΔH°).
Enthalpy and entropy of dissolution and solubility
Dissolution is thermodynamically favored when ΔG° = ΔH° − TΔS° is negative. An endothermic dissolution can be favored if TΔS° is larger than ΔH°, and an exothermic dissolution can be unfavored if ΔS° is sufficiently negative. At a given temperature, a more negative ΔG° of dissolution corresponds to a larger equilibrium constant for dissolution and, for salts of the same formula type, to a more soluble salt.
Cancellation among the contributions
The sign and relative size of each contribution to ΔG° of dissolution can be estimated, but the contributions have opposite signs and comparable sizes, so they largely cancel. Predicting the total ΔG° of dissolution, and so the solubility, from one factor alone is unreliable.
Students often think The hydrogen end of a water molecule is the negative end, so the H atoms of water point toward cations and the O atom points toward anions. In fact The oxygen end. The O atom of a water molecule carries a partial negative charge and is attracted to cations; the H atoms carry partial positive charges and are attracted to anions.
Students often think A dissolved ionic compound is present in the solution as intact units or molecules, such as MX or MgCl₂, each made of a cation joined to its anions. In fact No. A soluble ionic compound dissolves as separate cations and anions, each surrounded by water molecules. There are no molecules of the compound in the solid or in the solution.
4 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 4
The diagram shows a student's particle model of the dissolution of MgCl₂(s) in water. From the model, the student predicts that ΔS° for the dissolution is positive, but the measured ΔS° is negative. Which feature of the dissolution, missing from the student's model, best accounts for the negative ΔS°?
Answer and reasoning
AThe release of energy to the surroundings, which lowers the entropy of the solution A student who thinks the entropy change follows the direction of energy transfer picks this. The dissolution of MgCl₂ does release energy, but the sign of ΔS° is decided by the dispersal of the ions and the restriction of the water molecules, not by the sign of ΔH°.
BThe joining of Mg²⁺ and Cl⁻ ions into MgCl₂ molecules, which lowers the particle count A student who thinks a dissolved ionic compound exists as intact units picks this. MgCl₂ is present in solution as separate Mg²⁺ and Cl⁻ ions, as the model shows; the feature that the model omits is the water.
CThe disappearance of the ions as they dissolve, which leaves less matter to disperse A student who thinks dissolved particles are no longer there picks this. All the ions remain in the solution, dispersed among the water molecules; the entropy decrease comes from the water molecules that are held around the ions.
DThe ordered arrangement of water molecules around each ion, which restricts their motionCorrect The model draws only the ions, which become more dispersed, and leaves out the solvent. Water molecules are strongly attracted to the small, highly charged Mg²⁺ ion and to the Cl⁻ ions and are held in oriented positions around them, so the water becomes less free to move. For MgCl₂ this negative contribution to ΔS° outweighs the positive contribution from dispersing the ions.
For the dissolution of a certain ionic solid in water, ΔH° > 0 and ΔS° > 0, and both are approximately constant over the temperature range studied. How do ΔG° for the dissolution and Ksp of the solid change when the temperature is increased?
Answer and reasoning
AΔG° decreases, while Ksp decreases A student who thinks ΔG° and K rise and fall together picks this. ΔG° does decrease, but ln K = −ΔG°/RT = −ΔH°/RT + ΔS°/R, which increases with T when ΔH° > 0, so Ksp increases.
BΔG° is constant, and Ksp increases A student who treats ΔG° as a fixed standard value picks this, expecting only the solubility to respond to heating. ΔG° = ΔH° − TΔS° depends on T; with ΔS° > 0 it decreases as T increases.
CΔG° decreases, while Ksp increasesCorrect ΔG° = ΔH° − TΔS°. With ΔS° > 0, a larger T makes the term TΔS° larger, so ΔG° decreases. For the equilibrium constant, ln K = −ΔG°/RT = −ΔH°/RT + ΔS°/R; with ΔH° > 0 the term −ΔH°/RT becomes less negative as T increases, so Ksp increases and the solid becomes more soluble.
DΔG° increases, while Ksp decreases A student who treats the TΔS° term as added picks this: a larger T is taken to make ΔG° larger, and a larger ΔG° to mean a smaller Ksp. The term is subtracted: with ΔS° > 0, ΔG° decreases as T increases, and for an endothermic dissolution Ksp increases.
Working ΔG° = ΔH° − TΔS°. ΔH° and ΔS° are constant and ΔS° > 0, so as T increases, TΔS° increases and ΔG° decreases. K = e−ΔG°/RT = e−ΔH°/RT × eΔS°/R; with ΔH° > 0, −ΔH°/RT becomes less negative as T increases, so Ksp increases.
When NH₄NO₃(s) is stirred into water in an insulated cup, the solid dissolves quickly and the temperature of the solution decreases. NH₄NO₃ is very soluble in water at 25°C. Which reasoning correctly explains why the dissolution of NH₄NO₃ is thermodynamically favored at 25°C?
Answer and reasoning
AThe dissolution releases energy, as a favored process does, so ΔH° is negative A student who thinks a favored process must release energy picks this. The solution becomes colder, which shows that the dissolving solid absorbs energy from the water: ΔH° > 0. The process is favored because of its entropy increase.
BThe entropy increase is large enough that TΔS° is greater than ΔH°Correct The temperature decrease shows that the dissolution absorbs energy, so ΔH° > 0. The dissolution is nevertheless favored because ΔS° is positive and large enough that TΔS° exceeds ΔH°, which makes ΔG° = ΔH° − TΔS° negative.
CThe solid dissolves quickly, and a process with a high rate is favored A student who reads speed as evidence of favorability picks this. Rate and favorability are independent; the dissolution is favored because ΔG° = ΔH° − TΔS° is negative, not because it is fast.
DThe entropy of the system increases, and any process with ΔS° > 0 is favored A student who thinks an entropy increase of the system is enough picks this. ΔS° is positive here, but that alone does not decide favorability: the dissolution is favored only because TΔS° is larger than the positive ΔH°.
Solid MX contains M⁺ and X⁻ ions. Solid QZ contains Q²⁺ and Z²⁻ ions that are about the same sizes as M⁺ and X⁻. A student predicts that QZ is more soluble in water than MX because, according to Coulomb's law, the ions of QZ attract water molecules more strongly. Which evaluation of the student's prediction is correct?
Answer and reasoning
AUnreliable: the ions in solid QZ also attract one another more strongly, so the overall effect is uncertainCorrect The more highly charged ions do interact more strongly with water, which favors dissolving, but by the same law they attract one another more strongly in the solid, which opposes it, and they also restrict the surrounding water molecules more. These contributions to ΔG° of dissolution have opposite signs and largely cancel, so the solubility cannot be predicted from one factor.
BReliable: stronger attractions to water release more energy, and a more exothermic dissolution is more favored A student who thinks favorability is decided by the energy released picks this. The energy released by ion-water attractions is offset by the larger energy needed to separate the Q²⁺ and Z²⁻ ions, and the entropy contributions matter too, so the total ΔG° cannot be predicted this way.
CUnreliable: water molecules are neutral, so the charge of an ion has no effect on its attraction to water A student who treats water as a passive medium picks this. Water molecules are polar, and the ion-dipole attraction is stronger for more highly charged ions. The prediction is unreliable for a different reason: the attractions within the solid are stronger as well.
DUnreliable: the ions in solid QZ attract one another more strongly, so QZ must be the less soluble solid A student who thinks the attractions within the solid alone decide solubility picks this. The stronger attractions in solid QZ are opposed by stronger ion-water attractions, and these contributions largely cancel, so it is not possible to conclude that QZ is less soluble.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
