Solubility Equilibrium Practice Problems

Working Through Solubility Equilibrium Practice Problems

The hardest part about these problems isn't the math. It's knowing which equilibrium expression to write and whether you need to account for common ions, pH shifts, or complex formation before you even plug in numbers. I've spent years going through these with students, and the pattern is always the same: they rush to set up the Ksp equation without first checking what the problem is actually asking. Let me walk through how I approach them now. Start by writing out the dissolution equation. That means including state symbols. AgCl(s) Ag+(aq) + Cl-(aq). Don't skip the solid notation because it matters later when you're dealing with the common ion effect. The Ksp expression comes directly from that equation. For AgCl, Ksp = [Ag+][Cl-]. For something like PbCl2, it's Ksp = [Pb2+][Cl-]². The coefficient becomes the exponent. This seems obvious but I still see students writing [Cl-] without the square and then wondering why their answer is off by orders of magnitude. The real work starts when the problem gives you a solubility value and asks for Ksp, or vice versa. Here's where most people trip up. If you're given molar solubility 's' for a salt like Ca3(PO4)2, you need to think through the stoichiometry. Three calcium ions and two phosphate ions come out for every formula unit that dissolves. So [Ca2+] = 3s and [PO4³] = 2s. Then Ksp = (3s)³ × (2s)² = 108s. That fifth power is the kind of thing that eats into your calculation time if you're not comfortable with it.

I remember one student last semester who had a problem where the given solubility was 1.3 × 10³ mol/L for a hypothetical salt M2X3. They immediately wrote Ksp = s and got an answer that was nowhere near the expected value. The issue wasn't understanding Ksp conceptually. They just didn't work through the ion coefficients first. Once we went back to writing out [M³] = 2s and [X²] = 3s separately, the whole thing fell into place. That's the habit you want to build before touching a calculator.

When Q Tells You Whether Something Actually Precipitates

Not every solubility problem ends with finding a Ksp value. Sometimes you need to predict whether a precipitate forms when you mix two solutions. That's where the reaction quotient Q comes in. You calculate Q the same way you'd calculate Ksp, but using the actual concentrations at the moment of mixing, not the equilibrium concentrations. If Q > Ksp, precipitation occurs until Q drops to equal Ksp. If Q

Ksp, the solution is unsaturated and nothing precipitates. If Q = Ksp, you're exactly at equilibrium. The tricky part with mixing problems is remembering to recalculate concentrations after mixing. When you combine 50 mL of one solution with 50 mL of another, the total volume is 100 mL. Every ion concentration gets halved before you plug it into Q. I've lost count of the times I've seen people use the original concentrations and get a wildly wrong prediction. The dilution step is easy to forget under pressure. Here's a specific edge case that comes up more often than it should. Say you're mixing equal volumes of 0.01 M AgNO3 and 0.01 M NaCl. Both concentrations halve to 0.005 M after mixing. Q = (0.005)(0.005) = 2.5 × 10. Compare that to Ksp for AgCl which is 1.8 × 10¹. Q is enormously larger than Ksp. Precipitation will occur, and a significant amount of AgCl will fall out. But the final concentrations of Ag+ and Cl- in solution won't be zero. They'll both be approximately Ksp = 1.3 × 10 M. The amount remaining in solution is tiny compared to what started out, but it's not negligible if you're doing gravimetric work.

The Common Ion Effect and Why It Matters

When a salt dissolves in a solution that already contains one of its ions, solubility drops. This is the common ion effect and it shows up in practice problems constantly. The calculation is straightforward but students often miss the subtlety. If you're finding the solubility of AgCl in 0.1 M NaCl instead of pure water, you can't ignore the chloride already present. Set up the equation as Ksp = [Ag+][Cl-]total. The [Ag+] is your unknown 's' and the [Cl-]total is 0.1 + s. Since s is going to be very small compared to 0.1, you can approximate [Cl-] 0.1 and solve. s = Ksp / 0.1 = 1.8 × 10 M. In pure water, AgCl solubility is (1.8 × 10¹) = 1.3 × 10 M. The common ion reduces solubility by a factor of about 7,000. That's the kind of dramatic shift that makes this concept worth understanding, not just memorizing. The approximation step is where things can go wrong. You need to check that s is indeed negligible compared to the common ion concentration. In this case 1.8 × 10 is clearly much smaller than 0.1, so the approximation is valid. But if you ever get a result where s is more than 5% of the common ion concentration, you need to solve the full quadratic equation instead of approximating. I'd say roughly one in five practice problems tests whether you know when to make that check.

pH Effects on Solubility

This is where the problems get interesting. Salts containing basic anions become more soluble in acidic solution. The classic example is Mg(OH)2. In pure water its Ksp is about 1.8 × 10¹¹, giving a solubility around 1.7 × 10 M. But drop the pH and the hydroxide ions get protonated to form water. Le Chatelier's principle says the equilibrium shifts right to replace the consumed OH-, and more solid dissolves. For a problem asking for the solubility of Mg(OH)2 at pH 9, you first note that [OH-] = 10 M from the pH condition. Then Ksp = [Mg2+][OH-]² becomes 1.8 × 10¹¹ = s × (10)². Solving gives s = 180 M, which is obviously nonsensical and tells you the approximation broke down. What actually happens is the solid dissolves completely because the hydroxide concentration is being held so low by the buffer. The proper approach here is recognizing that at pH 9, the solubility is effectively unlimited for any reasonable amount of solid, and the limiting factor is how much you actually add. I ran into a genuinely subtle version of this once with a problem involving CaF2 in a buffered solution. The fluoride ion is a weak base, so it reacts with H+ to form HF. The equilibrium is CaF2(s) Ca2+ + 2F-, and then F- + H+ HF. You have to set up a system where both equilibria are satisfied simultaneously. The key insight is that the total fluorine in solution equals [F-] + [HF], and both depend on pH through the Ka of HF. The resulting equation for solubility involves the acid dissociation constant and the hydrogen ion concentration in a way that isn't immediately obvious. I usually recommend students write out all the relevant equilibria first, express everything in terms of one unknown, and then solve numerically or with successive approximation.

Miscibility Gaps and When Ksp Approaches Fail

Ksp values are useful but they have real limitations that practice problems rarely emphasize. The main issue is that Ksp assumes ideal behavior, meaning activity coefficients are all equal to one. In dilute solutions this is a reasonable assumption. In solutions with significant ionic strength, activities deviate from concentrations and your calculated solubilities can be off by a factor of two or three. For most introductory courses this isn't a concern, but if you're working with real data it's worth noting. Another limitation is that Ksp values are temperature-dependent. A Ksp listed in your textbook might be at 25°C while the problem is set at a different temperature. Without thermodynamic data you can't correct for this, so you just work with what's given. I've seen students lose points for questioning the temperature assumption when it's clearly implied by the problem context. Predicting solubility also breaks down for salts that form complex ions. Silver chloride dissolves in ammonia because [Ag(NH3)2]+ forms. The overall solubility depends on both Ksp and the formation constant Kf for the complex. You need to combine the two equilibria: AgCl(s) Ag+ + Cl- and Ag+ + 2NH3 [Ag(NH3)2]+. The net reaction is AgCl(s) + 2NH3 [Ag(NH3)2]+ + Cl- with an equilibrium constant of Ksp × Kf. This combined constant is what determines how much silver chloride dissolves in a given ammonia concentration. Problems that involve this are slightly more involved but the approach is systematic.

A Worked Example

Let me work through a complete example to show the thought process. Find the molar solubility of BaSO4 in a solution that already contains 0.01 M Na2SO4. Ksp for BaSO4 is 1.1 × 10¹. BaSO4(s) Ba2+(aq) + SO4²(aq). The Ksp expression is Ksp = [Ba2+][SO4²]. Let s be the molar solubility. Then [Ba2+] = s and [SO4²] = 0.01 + s from the sodium sulfate contribution plus the barium sulfate dissolution. Because Ksp is very small, s will be tiny compared to 0.01, so approximate [SO4²] 0.01. Then 1.1 × 10¹ = s × 0.01, giving s = 1.1 × 10 M. Check the approximation: 1.1 × 10 is indeed negligible compared to 0.01. The solubility in pure water would be (1.1 × 10¹) = 1.0 × 10 M. The common sulfate ion reduces solubility by about a thousandfold. That's the kind of problem that shows up repeatedly in different forms. The pattern recognition comes from doing enough of them that you stop second-guessing the setup and just work through the algebra. For Solubility Equilibrium Practice Problems, I'd recommend getting through at least twenty varied examples covering simple Ksp calculations, common ion effects, precipitation prediction, pH-dependent solubility, and complex ion formation. That covers the range you'll see on most exams and gives you enough pattern exposure to handle the variations they throw at you.

If you're looking for additional problems to work through, most general chemistry textbooks have dedicated sections with answers in the back. Online resources like Khan Academy and the Chemistry LibreTexts project also have problem sets with worked solutions. The key is doing the problems yourself first before looking at any solutions. Reading through someone else's work gives a false sense of confidence that disappears the moment you try it unaided. One final note on a practical detail that catches people out. When converting between solubility in g/L and molar solubility, make sure you're using the right molar mass. For a salt like PbI2, the molar mass is 461 g/mol, not the mass of lead or iodine alone. I've corrected this mistake in my own grading more times than I'd like to admit, and it's almost always a simple unit conversion error rather than a conceptual misunderstanding.