Solubility product

Drop silver chloride into water and only a whisper of it dissolves — the rest sits stubbornly as solid, in equilibrium with the tiny amount of Ag+Ag^+ and Cl−Cl^- that did escape into solution. That balance point has a name, solubility product KspK_{sp}, and it is the single number that lets a chemist predict, before mixing a single drop, whether two solutions will stay clear or throw down a precipitate. It closes out the ionic-equilibrium unit by extending the familiar KcK_c idea to a solid-liquid equilibrium, and it hands you two practical tools: converting between KspK_{sp} and molar solubility ss for salts of any ion ratio, and comparing an instantaneous ionic product QspQ_{sp} against KspK_{sp} to call a reaction's fate — unsaturated, saturated, or precipitating. The common-ion effect then shows the same suppression logic you already met with weak acids.

The core insight

For a sparingly soluble salt AxByA_xB_y, Ksp=[Ay+]x[Bx−]yK_{sp} = [A^{y+}]^x[B^{x-}]^y at equilibrium, and comparing the instantaneous ionic product QspQ_{sp} to this fixed KspK_{sp} tells you whether a solution is unsaturated, saturated, or about to precipitate.

The core idea

The equilibrium and its constant. A sparingly soluble ionic salt sits in a heterogeneous equilibrium with its own saturated solution:

AxBy(s)⇌xAy+(aq)+yBx−(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)

Because the solid has a fixed, constant activity (exactly as with any pure solid in a KcK_c expression), it drops out of the equilibrium expression, leaving only the dissolved ions:

Ksp=[Ay+]x[Bx−]yK_{sp} = [A^{y+}]^x[B^{x-}]^y

where: [Ay+][A^{y+}] and [Bx−][B^{x-}] are the equilibrium molar concentrations of the two ions, and xx, yy are the stoichiometric subscripts fixed by the salt's formula.

Linking KspK_{sp} to molar solubility. If the molar solubility of AxByA_xB_y is ss (mol L−1^{-1}), each formula unit that dissolves releases xx cations and yy anions, so [Ay+]=xs[A^{y+}] = xs and [Bx−]=ys[B^{x-}] = ys at equilibrium. Substituting gives the general conversion used throughout this topic, worked out fully in the derivation below.

QspQ_{sp} versus KspK_{sp} — predicting the outcome of mixing. For any solution, not necessarily at equilibrium, the ionic product QspQ_{sp} is computed with the same expression as KspK_{sp} but using whatever concentrations are actually present. Comparing the two numbers tells the whole story: Qsp<KspQ_{sp} < K_{sp} means unsaturated (more solid could dissolve), Qsp=KspQ_{sp} = K_{sp} means exactly saturated, and Qsp>KspQ_{sp} > K_{sp} means precipitation occurs until QspQ_{sp} falls back to KspK_{sp}.

The common-ion effect. Adding a soluble salt that supplies an ion already present (say, NaClNaCl added to a saturated AgClAgCl solution, boosting Cl−Cl^-) pushes the dissolution equilibrium backward by Le Chatelier's principle, so the sparingly soluble salt's solubility drops — the same suppression mechanism as a common ion silencing a weak acid's ionisation.

Figure
Three beakers showing unsaturated saturated and precipitating solutions compared against a fixed Ksp line
Comparing the ionic product Qsp against the fixed Ksp Ksp Qsp less than Ksp unsaturated Qsp equals Ksp saturated Qsp greater than Ksp precipitate forms
Derivation AFrom molar solubility to the solubility product

The goal is to turn a single measured quantity, the molar solubility ss of a sparingly soluble salt AxByA_xB_y, into its solubility product KspK_{sp}, and then to invert the same relation for a 1:2 salt so that the exponents on each ion are handled correctly rather than guessed, since this exponent bookkeeping is exactly where most numerical mistakes on this topic occur.

Step 1. Write the dissolution equilibrium and let the molar solubility be ss (mol L−1^{-1}): AxBy(s)⇌xAy+(aq)+yBx−(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq). Each mole of salt that dissolves releases xx moles of cation and yy moles of anion, so at equilibrium [Ay+]=xs[A^{y+}] = xs and [Bx−]=ys[B^{x-}] = ys.

Step 2. Substitute these into the solubility product expression: Ksp=[Ay+]x[Bx−]y=(xs)x(ys)yK_{sp} = [A^{y+}]^x[B^{x-}]^y = (xs)^x(ys)^y. Collecting the constants xxx^x and yyy^y separately from the variable part gives the general result:

Ksp=xx yy s(x+y)K_{sp} = x^x\,y^y\,s^{(x+y)}

Step 3. Apply this to a 1:1 salt such as AgClAgCl, where x=y=1x = y = 1: Ksp=11⋅11⋅s2=s2K_{sp} = 1^1 \cdot 1^1 \cdot s^{2} = s^2, so s=Ksps = \sqrt{K_{sp}} — the simplest case, ion concentration equals solubility directly.

Step 4. Apply it to a 1:2 salt such as PbI2PbI_2, where x=1x = 1, y=2y = 2: Ksp=11⋅22⋅s3=4s3K_{sp} = 1^1 \cdot 2^2 \cdot s^{3} = 4s^3, so s=Ksp/43s = \sqrt[3]{K_{sp}/4}. The factor of 44 comes purely from the anion's stoichiometric coefficient of 22 being squared, and it must never be dropped. Verified against the Pass-1 article: AgClAgCl, s=1.3×10−5 M⇒Ksp=s2=1.7×10−10s = 1.3\times10^{-5}\ M \Rightarrow K_{sp} = s^2 = 1.7\times10^{-10}; PbCl2PbCl_2, Ksp=1.6×10−5⇒s=Ksp/43=1.59×10−2 MK_{sp} = 1.6\times10^{-5} \Rightarrow s = \sqrt[3]{K_{sp}/4} = 1.59\times10^{-2}\ M.

Worked example

Common-ion suppression, then a precipitation call with QspQ_{sp}. Using Ksp(AgCl)=1.7×10−10K_{sp}(AgCl) = 1.7\times10^{-10} at 298 K (derived above from a pure-water solubility of 1.3×10−5 M1.3\times10^{-5}\ M): (a) find the molar solubility of AgClAgCl in a 0.10 M0.10\ M NaClNaCl solution, and (b) decide whether a precipitate forms when [Ag+]=1.0×10−6 M[Ag^+] = 1.0\times10^{-6}\ M is mixed with [Cl−]=1.0×10−3 M[Cl^-] = 1.0\times10^{-3}\ M.

Step 1. NaClNaCl is a strong electrolyte and dissociates fully, so before any AgClAgCl dissolves the solution already carries [Cl−]=0.10 M[Cl^-] = 0.10\ M. Let the extra solubility of AgClAgCl here be s′s': [Ag+]=s′[Ag^+] = s' and [Cl−]≈0.10+s′≈0.10 M[Cl^-] \approx 0.10 + s' \approx 0.10\ M, since the common ion suppresses s′s' down to a negligible size.

Step 2. Apply Ksp=[Ag+][Cl−]K_{sp} = [Ag^+][Cl^-]: 1.7×10−10=s′×0.101.7\times10^{-10} = s' \times 0.10, giving s′=1.7×10−100.10=1.7×10−9 mol L−1s' = \dfrac{1.7\times10^{-10}}{0.10} = 1.7\times10^{-9}\ \text{mol L}^{-1} — about 7600 times smaller than the 1.3×10−5 M1.3\times10^{-5}\ M solubility in pure water, confirming the common ion strongly suppresses dissolution and the approximation in Step 1 was safe to make.

Step 3. For part (b), compute the ionic product with the given concentrations: Qsp=[Ag+][Cl−]=(1.0×10−6)(1.0×10−3)=1.0×10−9Q_{sp} = [Ag^+][Cl^-] = (1.0\times10^{-6})(1.0\times10^{-3}) = 1.0\times10^{-9}.

Step 4. Compare Qsp=1.0×10−9Q_{sp} = 1.0\times10^{-9} with Ksp=1.7×10−10K_{sp} = 1.7\times10^{-10}: since Qsp>KspQ_{sp} > K_{sp}, the mixture is supersaturated with respect to AgClAgCl. A precipitate of AgClAgCl forms, and it continues to form until enough Ag+Ag^+ and Cl−Cl^- are removed from solution that QspQ_{sp} falls back down to KspK_{sp}.

Common traps

  • Comparing KspK_{sp} values directly across salts of different ion ratios. A numerically smaller KspK_{sp} does not automatically mean a less soluble salt if the stoichiometries differ — always convert both to molar solubility ss before comparing.
  • Dropping the stoichiometric exponents in the KspK_{sp} expression. A 1:2 salt gives Ksp=4s3K_{sp} = 4s^3, not s2s^2; every exponent must match the ion's coefficient in the balanced dissolution equation.
  • Mixing up KspK_{sp} with QspQ_{sp}. KspK_{sp} is a fixed equilibrium constant at a given temperature; QspQ_{sp} is computed the same way but with whatever instantaneous concentrations are actually present, and it is the QspQ_{sp}-versus-KspK_{sp} comparison that predicts precipitation.
  • Assuming the common ion's concentration is unaffected by the salt dissolving, without checking. The approximation [Cl−]≈ccommon[Cl^-] \approx c_{\text{common}} is only valid when suppression is strong; for very soluble salts the dissolved contribution can matter.

Ask the tutor: Want to try a precipitation-prediction problem that mixes two different salt solutions together, or see how this common-ion logic maps back onto buffer calculations?

Why does the solid not appear in the Ksp expression at all?

Because a pure solid has a fixed, constant activity that does not change as more or less of it is present — only the concentrations of species that can genuinely vary, the dissolved ions, appear in an equilibrium expression. This is the exact same rule you already use for any heterogeneous KcK_c involving a solid or pure liquid; KspK_{sp} is just that general rule applied to a dissolution equilibrium.

Can I compare two salts solubilities just by comparing their Ksp values?

Only if they have the same ion ratio (both 1:1, say, or both 1:2). Across different ratios the exponents differ, so a smaller KspK_{sp} can belong to the more soluble salt in mol L−1^{-1} terms. The safe method is always to convert each KspK_{sp} to its molar solubility ss using the salt's own stoichiometry, then compare the ss values directly.

How is the common-ion effect here related to what I learned about weak acids?

It is the identical Le Chatelier mechanism. Adding excess Cl−Cl^- from NaClNaCl shifts the AgClAgCl dissolution equilibrium backward, toward the solid, just as adding excess A−A^- shifts a weak acid HA⇌H++A−HA \rightleftharpoons H^+ + A^- backward and suppresses its ionisation. In both cases, raising the concentration of a product ion that is common to an added strong electrolyte pushes the equilibrium away from producing more of that ion.

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