1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
The hypothetical salt MX₂ dissolves according to the equation MX₂(s) ⇌ M²⁺(aq) + 2 X⁻(aq), with Ksp = 3.2 × 10⁻¹¹ at 25°C. What is the molar solubility of MX₂ at 25°C in a 0.020 M solution of NaX, a soluble salt?
Answer and reasoning
A2.0 × 10⁻⁴ M A student who thinks the solubility is the same in any solution picks this, the solubility in pure water from 4s³ = Ksp. The X⁻ already present raises the ion product, so much less MX₂ dissolves: s = Ksp/(0.020)² = 8.0 × 10⁻⁸ M.
B1.6 × 10⁻⁹ M A student who divides Ksp by [X⁻] without squaring it picks this: 3.2 × 10⁻¹¹/0.020 = 1.6 × 10⁻⁹ M. In Ksp = [M²⁺][X⁻]² the X⁻ term is squared, so s = 3.2 × 10⁻¹¹/(0.020)² = 8.0 × 10⁻⁸ M.
C2.0 × 10⁻⁸ M A student who doubles the X⁻ from NaX because of the coefficient 2 in the MX₂ equation picks this: 3.2 × 10⁻¹¹/(0.040)² = 2.0 × 10⁻⁸ M. NaX gives one X⁻ per formula unit, so [X⁻] = 0.020 M and s = 8.0 × 10⁻⁸ M.
D8.0 × 10⁻⁸ MCorrect NaX supplies [X⁻] = 0.020 M, far more than the 2s from MX₂, so Ksp = s(0.020)². Then s = (3.2 × 10⁻¹¹)/(4.0 × 10⁻⁴) = 8.0 × 10⁻⁸ M, much less than the 2.0 × 10⁻⁴ M solubility in pure water.
Working [X⁻] = 0.020 + 2s ≈ 0.020 M (NaX gives one X⁻ per formula unit). Ksp = [M²⁺][X⁻]² = s(0.020)² = 3.2 × 10⁻¹¹, so s = 3.2 × 10⁻¹¹/4.0 × 10⁻⁴ = 8.0 × 10⁻⁸ M. 2s = 1.6 × 10⁻⁷ M ≪ 0.020 M, so the approximation holds (the exact cubic gives the same two-figure value).
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.12.A.1 Common ion Fix
Common ion
An ion that is already present in a solution and is also one of the ions produced when a salt dissolves, such as Cl⁻ from NaCl(aq) for the dissolution of AgCl.
Common-ion effect
The reduction in the solubility of a salt when it is dissolved in a solution that already contains one of its ions. By Le Châtelier's principle, the added ion makes Q larger than Ksp, so the dissolution equilibrium lies further toward the solid.
Calculating solubility with a common ion present
The molar solubility in a solution containing a common ion is found from Ksp, with the concentration of the common ion included in the Ksp expression. When the common ion is much more concentrated than the amount the salt adds, its concentration can be used directly: for MX₂ in a solution with [X⁻] = c, s ≈ Ksp/c².
Ksp unchanged by a common ion
At a given temperature Ksp has one value. A common ion lowers the solubility, but the product of the ion concentrations, each raised to its power, at equilibrium still equals the same Ksp.
Precipitation from a saturated solution
If adding an ion makes the ion product Q greater than Ksp, the solution is no longer at equilibrium and solid forms until Q again equals Ksp.
Students often think Because Ksp is a constant at a given temperature, the molar solubility of a salt is the same whatever else is dissolved in the solution. In fact No. Ksp is constant at a given temperature, but the molar solubility depends on the other ions present: in a solution that already contains one of the salt's ions, less of the salt dissolves before the ion product reaches Ksp.
Students often think The concentration of the common ion is simply divided into Ksp, without being raised to the power of its coefficient. In fact Yes. For MX₂, Ksp = [M²⁺][X⁻]², so with [X⁻] fixed at c by a dissolved salt, s = Ksp/c².
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
A student measures the molar solubility of BaSO₄ by stirring excess solid BaSO₄ with 100 mL of distilled water at 25°C until equilibrium is reached and then analyzing the filtered solution. Which modification to the procedure would give a smaller measured molar solubility of BaSO₄ at 25°C?
Answer and reasoning
AUsing 100 mL of 0.010 M Na₂SO₄(aq) in place of the 100 mL of distilled waterCorrect Na₂SO₄ dissolves completely, supplying SO₄²⁻, an ion that appears in Ksp = [Ba²⁺][SO₄²⁻]. With SO₄²⁻ already present, the ion product reaches Ksp with much less BaSO₄ dissolved, so the measured molar solubility is smaller.
BUsing 100 mL of 0.010 M NaNO₃(aq) in place of the 100 mL of distilled water A student who thinks any dissolved salt lowers solubility like a common ion picks this. Neither Na⁺ nor NO₃⁻ appears in the Ksp expression for BaSO₄, so NaNO₃ does not cause a common-ion effect.
CUsing half as much excess solid BaSO₄, with some still undissolved A student who thinks less excess solid gives a less concentrated saturated solution picks this. As long as some solid remains undissolved, the solution is saturated and its concentration is set by Ksp.
DUsing 200 mL of distilled water with the same excess of solid BaSO₄ A student who thinks extra water dilutes a saturated solution picks this. More BaSO₄ dissolves in the larger volume, but with excess solid present the solution reaches the same concentration.
Working No calculation. BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). Na₂SO₄ supplies SO₄²⁻, a common ion, raising Q above Ksp, so less BaSO₄ dissolves. NaNO₃ shares no ion with BaSO₄; the amount of excess solid and the volume of water do not change the concentration of a saturated solution.
The hypothetical slightly soluble salt MX is dissolved at 25°C in a series of solutions of NaX, a soluble salt, of increasing concentration, from 0 M up to a concentration far greater than the molar solubility of MX in pure water. Which of the numbered graphs best shows how the molar solubility of MX, s, depends on the concentration of X⁻ supplied by the NaX?
Answer and reasoning
AGraph 1 A student who thinks a constant Ksp means a constant solubility picks the horizontal line. Ksp is constant, but with X⁻ already present the ion product reaches Ksp with less MX dissolved, so s falls as [X⁻] rises.
BGraph 2Correct Ksp = s(c + s) stays constant. At c = 0, s = √Ksp; as c rises, s must fall, and once c is much larger than s, s ≈ Ksp/c, so each doubling of c halves s. The curve falls steeply, then flattens, approaching zero without reaching it.
CGraph 3 A student who pictures the common-ion effect as removing a fixed amount of solubility per addition picks the straight line falling to zero. Because s ≈ Ksp/[X⁻], doubling [X⁻] halves s; s approaches zero but never reaches it.
DGraph 4 A student who takes the rising [X⁻] as a sign that more MX is dissolved picks the rising line. The X⁻ comes from NaX; s = [M⁺] = Ksp/[X⁻] falls as [X⁻] rises.
Working Ksp = [M⁺][X⁻] = s(c + s), where c = [X⁻] from NaX. At c = 0, s = √Ksp. As c increases, s = Ksp/(c + s) decreases; once c ≫ s, s ≈ Ksp/c, so doubling c halves s: a curve that falls steeply and then flattens, approaching but never reaching zero.
A clear, saturated solution of the slightly soluble salt MX, with no undissolved solid present, is at 25°C. A few drops of a concentrated solution of NaX, a soluble salt, are added and stirred in, and a solid precipitate forms. Which particulate-level explanation accounts for the precipitate?
Answer and reasoning
AThe saturated solution has no room for Na⁺ and X⁻ ions, so the added salt comes out as solid A student who thinks a saturated solution cannot dissolve any other solute picks this. The solution is saturated only with MX; NaX is soluble and dissolves, and the solid that forms is MX.
BThe added X⁻ makes Ksp for MX smaller, so some of the dissolved MX forms solid A student who thinks a common ion lowers Ksp picks this. Ksp for MX depends only on temperature; the added X⁻ raises the ion product above the unchanged Ksp, and that is why MX precipitates.
CThe added X⁻ makes [M⁺][X⁻] greater than Ksp, so M⁺ and X⁻ ions combine to form solid MXCorrect NaX dissociates completely, raising [X⁻]. The ion product [M⁺][X⁻] then exceeds Ksp, so M⁺ and X⁻ ions combine to form solid MX until the ion product falls back to Ksp.
DDissolved MX molecules are pushed out of the solution by the added NaX and settle as solid MX A student who pictures dissolved MX as intact molecules picks this. Dissolved MX is present as separate M⁺ and X⁻ ions; solid forms when these ions combine because the ion product exceeds Ksp.
Working No calculation. Before: [M⁺][X⁻] = Ksp. NaX dissociates completely, adding X⁻; now Q = [M⁺][X⁻] > Ksp, so M⁺ and X⁻ ions combine to form solid MX until Q = Ksp again. Ksp is unchanged at 25°C.
The table shows the measured molar solubility of the hypothetical salt MX at 25°C in pure water and in solutions of NaX, a soluble salt, of different concentrations. MX dissolves according to the equation MX(s) ⇌ M⁺(aq) + X⁻(aq). Which claim about Ksp for MX at 25°C is supported by the data?
Answer and reasoning
AKsp decreases as [NaX] increases, because the solubility of MX decreases A student who thinks a common ion works by lowering Ksp picks this. The solubility does decrease, but [M⁺][X⁻] is 1.0 × 10⁻¹⁰ in every solution, so Ksp is unchanged.
BKsp is the same in every solution, because [M⁺][X⁻] equals 1.0 × 10⁻¹⁰ in eachCorrect In pure water [M⁺][X⁻] = s² = (1.0 × 10⁻⁵)² = 1.0 × 10⁻¹⁰. In each NaX solution [X⁻] ≈ [NaX], and s × [NaX] is also 1.0 × 10⁻¹⁰ (for example 0.020 × 5.0 × 10⁻⁹). The solubility falls, but Ksp does not change.
CKsp increases as [NaX] increases, because the solutions contain more X⁻ A student who thinks adding one of the ions makes Ksp larger picks this. [X⁻] rises, but [M⁺] = s falls in proportion, so the product [M⁺][X⁻] stays at 1.0 × 10⁻¹⁰.
DKsp is the same in every solution, because it equals 1.0 × 10⁻⁵, the solubility in water A student who thinks Ksp is the molar solubility in pure water picks this. For MX, Ksp = s² = (1.0 × 10⁻⁵)² = 1.0 × 10⁻¹⁰, and the same value is found in every NaX solution.
Working Pure water: Ksp = s² = (1.0 × 10⁻⁵)² = 1.0 × 10⁻¹⁰. In NaX: [M⁺] = s, [X⁻] = [NaX] + s ≈ [NaX]: 0.010 × 1.0 × 10⁻⁸ = 1.0 × 10⁻¹⁰; 0.020 × 5.0 × 10⁻⁹ = 1.0 × 10⁻¹⁰; 0.050 × 2.0 × 10⁻⁹ = 1.0 × 10⁻¹⁰; 0.100 × 1.0 × 10⁻⁹ = 1.0 × 10⁻¹⁰. Ksp is the same, 1.0 × 10⁻¹⁰, in every solution.
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
