4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
Excess solid CaF₂ is stirred with water at 25°C until the solution is saturated. Which of the numbered diagrams best represents a small volume of the saturated solution above the solid? Water molecules are not shown, and the diagrams are not intended to show actual concentrations.
Answer and reasoning
ADiagram 1 A student who thinks the subscript 2 means F₂ units stay together picks the box with joined pairs of F⁻. Each formula unit that dissolves gives two separate F⁻ ions; there are no F₂ particles in the solution.
BDiagram 2Correct CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq). A small amount of the solid dissolves as separate ions, two F⁻ for every Ca²⁺, in equilibrium with the solid that remains. The box with two Ca²⁺ ions and four separate F⁻ ions shows this.
CDiagram 3 A student who thinks an ionic compound dissolves as intact CaF₂ units picks the box with joined Ca–F–F units. CaF₂ is an ionic solid; the part that dissolves separates into Ca²⁺ and F⁻ ions.
DDiagram 4 A student who thinks a slightly soluble salt does not dissolve at all picks the box with no dissolved particles. CaF₂ has a small Ksp, but a saturated solution still contains some Ca²⁺ and F⁻ ions.
Working No calculation. CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq): a small amount dissolves, giving separate Ca²⁺ and F⁻ ions in a 1 : 2 ratio, in equilibrium with the remaining solid. The correct box shows two Ca²⁺ ions and four separate F⁻ ions above the solid.
Calcium phosphate dissolves according to the equation Ca₃(PO₄)₂(s) ⇌ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq). If s is the molar solubility of Ca₃(PO₄)₂ in water, which relationship between Ksp and s is correct?
Answer and reasoning
AKsp = 72s⁵ A student who takes the exponents from the ionic charges picks this: (3s)²(2s)³ = 9s² × 8s³ = 72s⁵. The exponents are the coefficients in the dissolution equation, 3 for Ca²⁺ and 2 for PO₄³⁻, giving 108s⁵.
BKsp = 6s² A student who multiplies the ion concentrations without powers picks this: (3s)(2s) = 6s². Each concentration must be raised to the power of its coefficient, so Ksp = (3s)³(2s)² = 108s⁵.
CKsp = s² A student who uses the 1:1 relationship for every salt picks this. Ksp = s² holds only for a salt that gives one cation and one anion; Ca₃(PO₄)₂ gives five ions per formula unit, and Ksp = 108s⁵.
DKsp = 108s⁵Correct Each mole of Ca₃(PO₄)₂ that dissolves gives 3 mol Ca²⁺ and 2 mol PO₄³⁻, so [Ca²⁺] = 3s and [PO₄³⁻] = 2s. Ksp = [Ca²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 108s⁵.
Working [Ca²⁺] = 3s and [PO₄³⁻] = 2s. Ksp = [Ca²⁺]³[PO₄³⁻]² = (3s)³(2s)² = 27s³ × 4s² = 108s⁵.
The hypothetical salt AB dissolves according to the equation AB(s) ⇌ A⁺(aq) + B⁻(aq). At 25°C, Ksp for AB is 25. Which statement about AB at 25°C is correct?
Answer and reasoning
AIt is soluble, with a molar solubility of 5.0 MCorrect For a 1:1 salt, Ksp = [A⁺][B⁻] = s², so s = √25 = 5.0 M. A Ksp greater than 1 corresponds to a salt classified as soluble.
BIt is soluble, with a molar solubility of 25 M A student who thinks the molar solubility is the same number as Ksp picks this. Ksp = s² for AB, so s = √25 = 5.0 M.
CIt is slightly soluble, like all salts that have a Ksp A student who thinks only slightly soluble salts have a Ksp picks this. Every salt has a Ksp; its size decides the classification, and Ksp > 1 corresponds to a soluble salt.
DIt dissolves without limit, since its Ksp is above 1 A student who thinks K > 1 means dissolving goes to completion picks this. Dissolution of AB still reaches equilibrium: once [A⁺][B⁻] = 25, at s = 5.0 M, no more dissolves.
Working Ksp = s² = 25, so s = √25 = 5.0 M. Ksp > 1 corresponds to a soluble salt; the dissolution still reaches equilibrium once [A⁺][B⁻] = 25.
A student plans to determine Ksp for a slightly soluble salt, MX₂, by evaporating a measured volume of its filtered saturated solution to dryness and weighing the residue. Besides the volume evaporated and the mass of the residue, which quantity must the student know to calculate Ksp?
Answer and reasoning
AThe mass of solid MX₂ first added, to include it in Ksp A student who thinks the solid belongs in the Ksp expression picks this. Ksp contains only the dissolved ion concentrations, so the amount of solid added is not needed.
BThe volume of water first used, to find how much salt dissolved A student who thinks molar solubility depends on the volume of water used picks this. The concentration of a saturated solution is the same for any volume of water, and it is found from the measured volume that was evaporated.
CThe time taken to reach saturation, to find how fast it dissolved A student who links how fast a salt dissolves to how much can dissolve picks this. The rate of dissolving does not enter Ksp, which depends only on the ion concentrations at saturation.
DThe molar mass of MX₂, to convert the residue mass into molesCorrect The residue mass gives grams of MX₂ in the volume evaporated; the molar mass converts this to moles, giving s in mol/L, and then Ksp = s(2s)² = 4s³.
Working s (mol/L) = (mass of residue ÷ molar mass) ÷ volume evaporated (L); then Ksp = 4s³. The molar mass is the one extra quantity needed; the mass of solid added, the volume of water first used and the time taken do not enter.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.11.A.1 Solubility equilibrium Fix
Solubility equilibrium
The dynamic equilibrium in a saturated solution between an undissolved ionic solid and its dissolved ions, for example AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). Ions leave the solid and return to it at equal rates, so the concentrations of the dissolved ions stay constant.
Solubility-product constant, Ksp
The equilibrium constant for the dissolution of a salt: the product of the equilibrium concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced dissolution equation. The solid does not appear in the expression; for Ca₃(PO₄)₂, Ksp = [Ca²⁺]³[PO₄³⁻]².
Saturated solution
A solution in which the dissolved salt is at equilibrium with undissolved solid at a given temperature. Its ion concentrations do not depend on how much excess solid is present or on the volume of solution.
Students often think An ionic compound dissolves as intact molecules or formula units, so a solution of CaF₂ contains CaF₂ particles rather than separate Ca²⁺ and F⁻ ions. In fact Separate ions: each formula unit of CaF₂ that dissolves gives one Ca²⁺ ion and two F⁻ ions, which move independently, each surrounded by water molecules.
Students often think Atoms written together with a subscript stay together when the salt dissolves, so CaF₂ gives Ca²⁺ ions and F₂ units. In fact No. The subscript gives the ratio of ions in the solid. Each dissolved formula unit gives two separate F⁻ ions; there are no F₂ particles in the solution.
7.11.A.2 Molar solubility, s Fix
Molar solubility, s
The number of moles of a salt that dissolve per liter of solution to form a saturated solution at a given temperature. The ion concentrations follow from s and the formula: for MX₂, [M²⁺] = s and [X⁻] = 2s.
Relation between Ksp and molar solubility
Substituting the ion concentrations in terms of s into the Ksp expression gives Ksp as a function of s that depends on the formula: Ksp = s² for MX, 4s³ for MX₂ or M₂X, and 108s⁵ for M₃X₂.
Comparing solubilities with Ksp
Ksp values can be compared directly to rank molar solubilities only for salts that give the same number of ions in the same ratio (for example two MX salts). For salts of different formula types, calculate s for each salt and compare the values of s.
Students often think The molar solubility of any salt is the square root of Ksp, as if every salt dissolved to give one cation and one anion. In fact No. s = √Ksp holds only for a salt that gives one cation and one anion (MX). For other formulas the ion concentrations are multiples of s, so for MX₂, Ksp = s(2s)² = 4s³.
Students often think The molar solubility of a salt is equal to its Ksp, since both describe how much of the salt dissolves. In fact No. Ksp is a product of ion concentrations raised to powers, while s is the moles of salt dissolved per liter; for MX, s = √Ksp, and for MX₂, s = (Ksp/4)1/3.
7.11.A.3 Ksp and the solubility rules Fix
Ksp and the solubility rules
The solubility rules can be related to Ksp: a Ksp value greater than 1 corresponds to a salt classified as soluble (such as the common sodium, potassium, ammonium and nitrate salts), whereas salts described as insoluble or slightly soluble have Ksp values much less than 1.
Students often think Only slightly soluble salts have a Ksp, so any salt for which a Ksp is given must be slightly soluble. In fact No. Every salt has a Ksp: the dissolution of the solid in contact with its saturated solution is an equilibrium. Very soluble salts have Ksp values greater than 1; slightly soluble salts have Ksp values much less than 1.
Students often think If Ksp is greater than 1, dissolving goes to completion, so any amount of the salt dissolves in any amount of water. In fact No. A Ksp greater than 1 means that the salt is soluble and the saturated solution is concentrated, but the dissolution still reaches equilibrium: once the ion product equals Ksp, no more salt dissolves.
7.11.A.4 Determining Ksp from measurements Fix
Determining Ksp from measurements
Ksp can be calculated from a measured molar solubility (for example from the mass of salt in a known volume of saturated solution and its molar mass) or from a measured concentration of one ion in the saturated solution, using the formula of the salt to find the concentrations of all the ions.
Students often think The molar solubility depends on the volume of water used, because more water dissolves more of the salt, so the volume used to make the saturated mixture is needed to find s and Ksp. In fact No. With excess solid present, a larger volume of water dissolves a larger amount of salt, but the concentration of the saturated solution, and so the molar solubility, is the same.
Students often think The concentration of an ion must be multiplied by its coefficient and then raised to the power of that coefficient, even when the ion concentration itself has been measured, so for M₃X the Ksp is (3[M⁺])³[X³⁻]. In fact No. The Ksp expression uses the actual ion concentrations, each raised to the power of its coefficient. A factor such as 3 in (3s)³ only converts the molar solubility s into the ion concentration; a measured ion concentration already includes it.
7 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 7
The hypothetical salt MX₂ dissolves according to the equation MX₂(s) ⇌ M²⁺(aq) + 2 X⁻(aq). At 25°C, Ksp for MX₂ is 4.0 × 10⁻⁹. What is the molar solubility of MX₂ in water at 25°C?
Answer and reasoning
A1.0 × 10⁻³ MCorrect With [M²⁺] = s and [X⁻] = 2s, Ksp = s(2s)² = 4s³. So s³ = (4.0 × 10⁻⁹)/4 = 1.0 × 10⁻⁹ and s = 1.0 × 10⁻³ M.
B6.3 × 10⁻⁵ M A student who uses s = √Ksp for every salt picks this: √(4.0 × 10⁻⁹) = 6.3 × 10⁻⁵ M. MX₂ gives three ions per formula unit, so Ksp = 4s³ and s = 1.0 × 10⁻³ M.
C1.6 × 10⁻³ M A student who sets [X⁻] = s instead of 2s picks this: Ksp = s × s² = s³, so s = (4.0 × 10⁻⁹)1/3 = 1.6 × 10⁻³ M. Each MX₂ that dissolves gives two X⁻ ions, so [X⁻] = 2s and Ksp = 4s³.
D4.0 × 10⁻⁹ M A student who thinks the molar solubility is the same number as Ksp picks this. Ksp is a product of ion concentrations raised to powers; here Ksp = 4s³, so s = 1.0 × 10⁻³ M.
Working Let s = molar solubility: [M²⁺] = s, [X⁻] = 2s. Ksp = s(2s)² = 4s³ = 4.0 × 10⁻⁹, so s³ = 1.0 × 10⁻⁹ and s = 1.0 × 10⁻³ M.
The hypothetical salt M₂X dissolves according to the equation M₂X(s) ⇌ 2 M⁺(aq) + X²⁻(aq). In a saturated solution of M₂X at 25°C, the concentration of M⁺ is measured to be 6.0 × 10⁻³ M. What is the value of Ksp for M₂X at 25°C?
Answer and reasoning
A2.2 × 10⁻⁷ A student who thinks every ion of the salt has the same concentration picks this, using [X²⁻] = 6.0 × 10⁻³ M: (6.0 × 10⁻³)³ = 2.2 × 10⁻⁷. Each formula unit gives two M⁺ but only one X²⁻, so [X²⁻] = 3.0 × 10⁻³ M.
B1.8 × 10⁻⁵ A student who writes Ksp without powers picks this: [M⁺][X²⁻] = (6.0 × 10⁻³)(3.0 × 10⁻³) = 1.8 × 10⁻⁵. The coefficient 2 for M⁺ makes the expression [M⁺]²[X²⁻] = 1.1 × 10⁻⁷.
C1.1 × 10⁻⁷Correct Each X²⁻ ion is accompanied by two M⁺ ions, so [X²⁻] = ½(6.0 × 10⁻³ M) = 3.0 × 10⁻³ M. Ksp = [M⁺]²[X²⁻] = (6.0 × 10⁻³)²(3.0 × 10⁻³) = 1.1 × 10⁻⁷.
D9.0 × 10⁻⁶ A student who treats every salt as 1:1, with Ksp = s², picks this: (3.0 × 10⁻³)² = 9.0 × 10⁻⁶. M₂X gives three ions per formula unit, so Ksp = [M⁺]²[X²⁻] = (2s)²s = 1.1 × 10⁻⁷.
Working [M⁺] = 6.0 × 10⁻³ M, so the molar solubility s = [X²⁻] = ½[M⁺] = 3.0 × 10⁻³ M. Ksp = [M⁺]²[X²⁻] = (6.0 × 10⁻³)²(3.0 × 10⁻³) = 1.08 × 10⁻⁷ = 1.1 × 10⁻⁷.
The table gives the Ksp values of two silver salts at 25°C. A student claims that AgCl is more soluble in water than Ag₂CrO₄ because AgCl has the larger Ksp. Which evaluation of the student's claim is correct? (s is the molar solubility.)
Answer and reasoning
ACorrect: the salt with the larger Ksp has the greater molar solubility A student who compares Ksp values directly for any two salts picks this. AgCl gives two ions and Ag₂CrO₄ gives three, so s enters their Ksp expressions differently; calculating s shows Ag₂CrO₄ (6.5 × 10⁻⁵ M) is more soluble than AgCl (1.3 × 10⁻⁵ M).
BIncorrect: 4s³ = Ksp gives s = 6.5 × 10⁻⁵ M for Ag₂CrO₄, more than for AgClCorrect For Ag₂CrO₄, [Ag⁺] = 2s and [CrO₄²⁻] = s, so Ksp = 4s³ and s = (1.1 × 10⁻¹²/4)1/3 = 6.5 × 10⁻⁵ M. For AgCl, s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. Ag₂CrO₄ is the more soluble salt, even though its Ksp is smaller.
CCorrect: s = √Ksp gives s = 1.0 × 10⁻⁶ M for Ag₂CrO₄, less than for AgCl A student who uses s = √Ksp for every salt picks this. Ag₂CrO₄ gives two Ag⁺ and one CrO₄²⁻, so Ksp = 4s³ and s = 6.5 × 10⁻⁵ M, more than the 1.3 × 10⁻⁵ M of AgCl.
DIncorrect: s³ = Ksp gives s = 1.0 × 10⁻⁴ M for Ag₂CrO₄, more than for AgCl A student who sets [Ag⁺] = s instead of 2s picks this, writing Ksp = s² × s = s³. The verdict is right but the calculation is not: [Ag⁺] = 2s, so Ksp = 4s³ and s = 6.5 × 10⁻⁵ M.
Working AgCl: Ksp = s², s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. Ag₂CrO₄ ⇌ 2 Ag⁺ + CrO₄²⁻: Ksp = (2s)²s = 4s³, s = (1.1 × 10⁻¹²/4)1/3 = 6.5 × 10⁻⁵ M. Ag₂CrO₄ has the greater molar solubility, so the claim is incorrect.
A saturated solution of AgCl is in contact with excess solid AgCl at 25°C. A small amount of solid AgCl made with radioactive silver, which behaves chemically like ordinary silver, is added and the mixture is stirred at 25°C. Filtered samples of the solution are analyzed at intervals. The table shows the results; radioactivity is given above the background level. Which particulate-level explanation is consistent with both sets of measurements?
Answer and reasoning
AAg⁺ ions stopped leaving the solid at saturation, so no exchange takes place A student who thinks dissolving stops at saturation picks this. The rising radioactivity of the solution shows that Ag⁺ ions from the added solid are still entering the solution.
BThe added solid made more AgCl dissolve, so the solution now holds more Ag⁺ ions A student who thinks adding solid to a saturated solution raises the ion concentrations picks this. The table shows [Ag⁺] unchanged at 1.3 × 10⁻⁵ M; the solid does not appear in Ksp, so adding it does not change the saturated concentration.
CAg⁺ ions leave the solid and rejoin it at equal rates, so [Ag⁺] stays constantCorrect Radioactive Ag⁺ appears in the solution, so ions are still leaving the solid; yet [Ag⁺] stays at 1.3 × 10⁻⁵ M, so ions must be rejoining the solid at the same rate. Dissolving and precipitation continue at equal rates: a dynamic equilibrium.
DAg⁺ ions leave the solid but none rejoin it, so the solid slowly dissolves A student who thinks dissolving is one-way picks this. If ions only left the solid, [Ag⁺] would keep rising, but the table shows it constant at 1.3 × 10⁻⁵ M, so ions must also be rejoining the solid.
Working No calculation. [Ag⁺] stays at 1.3 × 10⁻⁵ M (= √Ksp for AgCl), so there is no net dissolving. Radioactive Ag⁺ appears in the solution, so Ag⁺ ions are still leaving the solid; since [Ag⁺] does not rise, an equal number of Ag⁺ ions must be rejoining the solid. Dynamic equilibrium: equal rates of dissolving and precipitation.
To determine Ksp for a slightly soluble salt, a student stirs an excess of the solid with 100 mL of water at 25°C for 30 minutes, filters the mixture, evaporates 25.00 mL of the filtrate to dryness and weighs the residue. The student's Ksp is greater than the accepted value at 25°C. Which of the following, if it happened, could explain the high value?
Answer and reasoning
ATwice the intended mass of excess solid was stirred with the water A student who thinks more solid makes more of the salt dissolve picks this. Once excess solid is present, the solution is saturated and its concentration does not depend on how much solid remains.
BThe same excess of solid was stirred with 200 mL of water, not 100 mL A student who thinks the molar solubility depends on the volume of water picks this. More salt dissolves in 200 mL, but the concentration of the saturated solution, and so the residue from 25.00 mL, is the same.
CThe mixture was stirred for 60 minutes instead of 30 minutes A student who thinks longer stirring dissolves more salt even at saturation picks this. Stirring longer can only bring the solution closer to saturation; it cannot raise the concentration above the saturated value.
DFine particles of undissolved solid passed through the filter paperCorrect Undissolved particles in the filtrate are left in the residue along with the dissolved salt, so the residue mass, the calculated molar solubility and the calculated Ksp are all too high.
Working No calculation. Ksp is calculated from the mass of residue in 25.00 mL of filtrate. Undissolved particles passing through the filter add to the residue mass, so the calculated molar solubility and Ksp are too high. More excess solid, more water with excess solid still present, or longer stirring cannot raise the concentration of the saturated solution.
A student evaporated different volumes of a filtered saturated solution of the hypothetical salt MX₂ (molar mass 100. g/mol) at 25°C and weighed the dry residue each time. The graph shows the results with the best-fit line. MX₂ dissolves according to the equation MX₂(s) ⇌ M²⁺(aq) + 2 X⁻(aq). Based on the graph, what is Ksp for MX₂ at 25°C?
Answer and reasoning
A1.1 × 10⁻⁴Correct The slope of the line is 0.300 g per 100 mL, 3.00 g/L, so the molar solubility is s = 3.00/100. = 0.0300 M. For MX₂, [M²⁺] = s and [X⁻] = 2s, so Ksp = 4s³ = 4(0.0300)³ = 1.1 × 10⁻⁴.
B9.0 × 10⁻⁴ A student who uses Ksp = s² for every salt picks this: (0.0300)² = 9.0 × 10⁻⁴. MX₂ gives three ions per formula unit, so Ksp = s(2s)² = 4s³ = 1.1 × 10⁻⁴.
C2.7 × 10⁻⁵ A student who takes [X⁻] equal to s picks this: s × s² = (0.0300)³ = 2.7 × 10⁻⁵. Each MX₂ that dissolves gives two X⁻ ions, so [X⁻] = 2s and Ksp = 4s³ = 1.1 × 10⁻⁴.
D3.0 × 10⁻² A student who thinks Ksp is the same number as the molar solubility picks this, reporting s = 0.0300 M as Ksp. Ksp is the product of the ion concentrations raised to powers: 4s³ = 1.1 × 10⁻⁴.
Excess solid of the hypothetical salt M₃X is stirred with water at 25°C, and the concentrations of its ions in the solution are measured as the salt dissolves according to the equation M₃X(s) ⇌ 3 M⁺(aq) + X³⁻(aq). The graph shows the results. Based on the graph, what is Ksp for M₃X at 25°C?
Answer and reasoning
A3.0 × 10⁻⁴ A student who writes Ksp as the product of the ion concentrations without powers picks this: (0.030)(0.010) = 3.0 × 10⁻⁴. The coefficient 3 of M⁺ is an exponent, so Ksp = (0.030)³(0.010) = 2.7 × 10⁻⁷.
B3.0 × 10⁻⁸ A student who takes the exponents from the ionic charges picks this: [M⁺]¹[X³⁻]³ = (0.030)(0.010)³ = 3.0 × 10⁻⁸. The exponents are the coefficients in the dissolution equation, so Ksp = [M⁺]³[X³⁻] = 2.7 × 10⁻⁷.
C2.7 × 10⁻⁷Correct The flat parts of the curves give the ion concentrations in the saturated solution: [M⁺] = 0.030 M and [X³⁻] = 0.010 M. Ksp = [M⁺]³[X³⁻] = (0.030)³(0.010) = 2.7 × 10⁻⁷.
D7.3 × 10⁻⁶ A student who multiplies the measured [M⁺] by its coefficient before cubing it picks this: (3 × 0.030)³(0.010) = 7.3 × 10⁻⁶. The graph already gives the concentration of M⁺ itself, so it is used directly: (0.030)³(0.010) = 2.7 × 10⁻⁷.
Working The solution is saturated where the curves are flat: [M⁺] = 0.030 M and [X³⁻] = 0.010 M. Ksp = [M⁺]³[X³⁻] = (0.030)³(0.010) = (2.7 × 10⁻⁵)(0.010) = 2.7 × 10⁻⁷. Distractors: no powers, (0.030)(0.010) = 3.0 × 10⁻⁴; charges used as exponents, (0.030)(0.010)³ = 3.0 × 10⁻⁸; measured [M⁺] multiplied by 3 again, (3 × 0.030)³(0.010) = 7.3 × 10⁻⁶.
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
