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Free MCAT-style practice, 5A

MCAT Solubility, Ksp and Common Ion Practice Questions

By StudyStore · Published 2026-11-01 · 9 original questions · about 14 min

Nine free MCAT-style practice questions on solubility rules, Ksp and molar solubility, Q versus Ksp, the common ion effect and complex ions, with full explanations.

Quick concept primer

Ksp is the equilibrium constant for a sparingly soluble salt dissolving into its ions, written without the solid. To get molar solubility s, write each ion concentration in terms of s using the formula's coefficients: a 1:1 salt gives Ksp = s2, a 1:2 or 2:1 salt gives Ksp = 4s3. Because the exponents differ, a larger Ksp only means a larger molar solubility when the salts have the same ion ratio.

To predict precipitation, compute the ion product Q from the concentrations actually present after mixing. If Q is greater than Ksp, solid forms until Q falls back to Ksp; if Q is less, everything stays dissolved. Adding an ion the salt already contains (a common ion) pushes the equilibrium toward the solid and lowers solubility. Removing an ion raises solubility: acid removes basic anions such as hydroxide, and a ligand that binds the metal as a complex ion removes free metal ions.

For quick precipitation calls, nitrates, group 1 and ammonium salts are soluble, while many iodides, sulfides and carbonates of heavier metals are not. The 5A Water, Solutions, Acids and Bases pack adds flashcards and more practice on solubility, Ksp, acids, bases and buffers: get the 5A pack, or try the free sampler first.

Traps: dilute before computing Q when two solutions are mixed, and remember that Ksp itself changes only with temperature, not with added ions.

Free notes for this topic: Colligative Properties: Freezing Point Depression Calculations

9 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A sparingly soluble salt with the formula QY2 dissolves as QY2(s) → Q2+ + 2 Y-. Its molar solubility in pure water at 25 C is 3.0 × 10-4 M. What is its Ksp?

  1. 2.7 × 10-11
  2. 5.4 × 10-11
  3. 1.1 × 10-10
  4. 9.0 × 10-8
Show answer and explanation

Answer: C. 1.1 × 10-10

[Q2+] = s and [Y-] = 2s, so Ksp = s(2s)2 = 4s3 = 4(3.0 × 10-4)3 = 1.08 × 10-10. Choice A is s3 with the factor of 4 forgotten, choice B uses s2 × 2s instead of s × (2s)2, and choice D treats the salt as 1:1 (s2).

Question 2

A 1:1 salt MZ has Ksp = 1.6 × 10-9 at 25 C. What is its molar solubility in a 0.010 M solution of the fully soluble salt NaZ?

  1. 1.6 × 10-7 M
  2. 4.0 × 10-5 M
  3. 1.6 × 10-11 M
  4. 4.0 × 10-3 M
Show answer and explanation

Answer: A. 1.6 × 10-7 M

The 0.010 M Z- from NaZ is a common ion. With s much smaller than 0.010, Ksp = s(0.010), so s = 1.6 × 10-9 / 0.010 = 1.6 × 10-7 M. Choice B is the solubility in pure water, which ignores the common ion. Choice C multiplies by 0.010 instead of dividing, and choice D is not supported by any setup.

Question 3

A student tabulates three sparingly soluble salts at 25 C:

SaltCation : anion ratioKsp
P1 : 11.0 × 10-10
R2 : 13.2 × 10-14
T1 : 24.0 × 10-12

Which salt has the greatest molar solubility in pure water?

  1. Salt P, because it has the largest Ksp
  2. Salt R, because it releases the most cations per formula unit
  3. All three are equal, because each Ksp is very small
  4. Salt T, because its molar solubility works out highest once the ion ratio is used
Show answer and explanation

Answer: D. Salt T, because its molar solubility works out highest once the ion ratio is used

For P, s = (1.0 × 10-10)1/2 = 1.0 × 10-5 M. For R and T, Ksp = 4s3, giving s = 2.0 × 10-5 M for R and 1.0 × 10-4 M for T. T is the most soluble even though P has the largest Ksp; Ksp values can only be compared directly when the ion ratios match.

Question 4

At 25 C a 1:1 salt XW has Ksp = 1.0 × 10-8. A chemist mixes 50 mL of 4.0 × 10-4 M X(NO3) with 50 mL of 4.0 × 10-4 M NaW. What happens?

  1. No precipitate forms, because Q = 1.6 × 10-7 is below Ksp
  2. A precipitate forms, because Q = 4.0 × 10-8 exceeds Ksp
  3. No precipitate forms, because Q = 4.0 × 10-8 equals Ksp after dilution
  4. A precipitate forms, because Q = 1.6 × 10-7 exceeds Ksp
Show answer and explanation

Answer: B. A precipitate forms, because Q = 4.0 × 10-8 exceeds Ksp

Mixing equal volumes halves each concentration, so [X+] = [W-] = 2.0 × 10-4 M and Q = 4.0 × 10-8, which is four times Ksp, so solid forms. Choice D reaches the right outcome with undiluted concentrations, choice A compares the numbers in the wrong direction, and choice C misstates Q relative to Ksp.

Question 5

A metal hydroxide M(OH)2 dissolves endothermically. Each of the following would increase its molar solubility EXCEPT:

  1. adding hydrochloric acid to the suspension
  2. adding a ligand that binds M2+ as a soluble complex ion
  3. adding sodium hydroxide to the suspension
  4. warming the suspension
Show answer and explanation

Answer: C. adding sodium hydroxide to the suspension

NaOH supplies hydroxide, a common ion, which shifts the equilibrium toward the solid and lowers solubility. Acid removes OH- as water and the ligand removes free M2+, so both pull the dissolution forward. For an endothermic process, heating raises Ksp and therefore raises solubility.

Question 6

Solid silver chloride barely dissolves in water but dissolves noticeably in aqueous ammonia. Which statement best explains this?

  1. Ammonia binds Ag+ as a complex ion, lowering free [Ag+] so that Q drops below Ksp and more solid dissolves
  2. Ammonia raises the Ksp of silver chloride at constant temperature
  3. Ammonia is a common ion that shifts the equilibrium toward the dissolved ions
  4. Ammonia oxidizes chloride ions, removing them from solution
Show answer and explanation

Answer: A. Ammonia binds Ag+ as a complex ion, lowering free [Ag+] so that Q drops below Ksp and more solid dissolves

Silver ion and ammonia form the complex Ag(NH3)2+, which removes free Ag+. The ion product falls below Ksp and more AgCl dissolves. Ksp depends only on temperature (B is wrong), ammonia is not one of the salt's ions (C), and ammonia does not oxidize chloride (D).

Question 7

Equal volumes of 0.10 M aqueous solutions are mixed in pairs. Which pair produces a precipitate?

  1. Sodium nitrate and potassium chloride
  2. Ammonium chloride and potassium nitrate
  3. Sodium sulfate and potassium chloride
  4. Lead(II) nitrate and sodium iodide
Show answer and explanation

Answer: D. Lead(II) nitrate and sodium iodide

Lead(II) iodide is poorly soluble and precipitates as a yellow solid when Pb2+ meets I-. In choices A, B and C every possible product pairs a group 1 or ammonium cation with a nitrate, chloride or sulfate anion, and all of those salts are soluble, so nothing forms.

Question 8

A salt AB2 has Ksp = 4.0 × 10-12 at 25 C. What is its approximate molar solubility in 0.10 M NaB, where NaB is fully soluble?

  1. 1.0 × 10-4 M
  2. 4.0 × 10-10 M
  3. 4.0 × 10-11 M
  4. 2.0 × 10-6 M
Show answer and explanation

Answer: B. 4.0 × 10-10 M

With [B-] fixed near 0.10 M by the common ion, Ksp = s(0.10)2, so s = 4.0 × 10-12 / 0.010 = 4.0 × 10-10 M. Choice A is the pure-water solubility, and choice C forgets to square the common-ion concentration. Choice D does not follow from any consistent setup.

Question 9

A saturated aqueous solution of the salt G2H, which dissolves as G2H(s) → 2 G+ + H2-, contains [G+] = 4.0 × 10-4 M at 25 C. What is the Ksp of G2H?

  1. 3.2 × 10-11
  2. 2.6 × 10-10
  3. 1.6 × 10-7
  4. 1.6 × 10-11
Show answer and explanation

Answer: A. 3.2 × 10-11

Each formula unit gives two G+ and one H2-, so [H2-] = 2.0 × 10-4 M and Ksp = (4.0 × 10-4)2(2.0 × 10-4) = 3.2 × 10-11. Choice B wrongly treats 4.0 × 10-4 M as the molar solubility, and choice C squares [G+] but leaves out the anion term.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.

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