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AP Chemistry Notes

7.11.4 Using molar solubility to find Ksp

AP Syllabus focus: ‘Measured molar solubility (ion concentrations) in a saturated solution can be used to calculate the Ksp of a slightly soluble ionic compound.’

Slightly soluble ionic solids establish an equilibrium between undissolved solid and dissolved ions. If you can measure the ion concentrations in a saturated solution, you can determine KspK_{sp} quantitatively from stoichiometry.

A saturated solution contains the maximum dissolved amount of a salt at a given temperature, with undissolved solid present.

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Sequence of beakers illustrating a sparingly soluble salt as more solid is added: the solution concentration increases until the saturation point is reached. Beyond saturation, additional solid accumulates at the bottom while the dissolved ion concentrations remain constant—evidence of equilibrium between solid and ions in solution. Source

This ensures the dissolution reaction is truly at equilibrium, so the equilibrium-constant expression applies.

Key term used in calculations

Molar solubility (ss): the number of moles of an ionic compound that dissolve per litre of solution to form a saturated solution (units: mol L1\text{mol L}^{-1}).

Molar solubility is often not measured directly; instead, you may measure equilibrium ion concentrations (for example, via conductivity, titration, or spectroscopy) and relate them to ss using the balanced dissolution equation.

Writing the dissolution reaction correctly

To calculate KspK_{sp}, first write the balanced dissolution equilibrium for the salt. For a generic salt MaXbM_aX_b:

  • Include the solid as a reactant: MaXb(s)M_aX_b(s)

  • Write aqueous ions with correct stoichiometric coefficients and charges

  • Ensure the equation is balanced for both atoms and charge

The stoichiometric coefficients determine how ion concentrations depend on ss (or on each other if ss is not explicitly used).

From measured ion concentrations to KspK_{sp}

For slightly soluble salts, AP Chemistry typically treats dilute aqueous solutions as ideal enough that concentrations can be used in place of activities. Solids are omitted from the equilibrium expression.

Ksp=[Mn+]a[Xm]b K_{sp} = [M^{n+}]^{a}[X^{m-}]^{b}

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Plot showing combinations of ion concentrations that satisfy the equilibrium condition for a sparingly soluble salt (example shown for a 1:1 salt like AgCl). Points on the curve represent states where the ion product equals KspK_{sp}; points off the curve correspond to Q<KspQ<K_{sp} (unsaturated) or Q>KspQ>K_{sp} (supersaturated/precipitation until equilibrium is restored). Source

KspK_{sp} = solubility-product constant (unitless in AP practice)

[Mn+][M^{n+}] = equilibrium molar concentration of the cation, mol L1\text{mol L}^{-1}

[Xm][X^{m-}] = equilibrium molar concentration of the anion, mol L1\text{mol L}^{-1}

a,ba, b = stoichiometric coefficients from MaXb(s)aMn++bXmM_aX_b(s) \rightleftharpoons aM^{n+} + bX^{m-}

To use measured data, connect the measured equilibrium concentrations to the dissolution stoichiometry:

  • If MaXb(s)aMn++bXmM_aX_b(s) \rightleftharpoons aM^{n+}+bX^{m-} and the molar solubility is ss, then ideally:

    • [Mn+]=as[M^{n+}] = a s

    • [Xm]=bs[X^{m-}] = b s

  • If ion concentrations are measured directly, you may not need ss at all; substitute the measured equilibrium ion concentrations into the KspK_{sp} expression.

  • If only one ion concentration is measured:

    • Use stoichiometric ratios to infer the other ion concentration (e.g., if [Mn+][M^{n+}] is known, then [Xm]=ba[Mn+][X^{m-}] = \frac{b}{a}[M^{n+}]).

Practical checklist for “find KspK_{sp} from molar solubility”

  • Confirm the solution is saturated (equilibrium established with excess solid).

  • Write the balanced dissolution equation.

  • Translate molar solubility (ss) into equilibrium ion concentrations using coefficients.

  • Substitute equilibrium ion concentrations into the KspK_{sp} expression.

  • Report KspK_{sp} to a reasonable number of significant figures consistent with the measured data.

FAQ

Because $K_{sp}$ applies at equilibrium for the dissolution reaction. Undissolved solid confirms saturation, meaning ion concentrations are equilibrium values rather than just “some dissolved amount” below saturation.

In more concentrated ionic solutions, electrostatic interactions make activities smaller than concentrations. Using $[,]$ can then give an “apparent” $K_{sp}$. A more rigorous approach uses activity coefficients $\gamma$ so $a_i=\gamma_i[i]$.

Common approaches include:

  • Titration (e.g., precipitating an ion with a standard solution)

  • Conductivity (relating total ion concentration to measured conductance)

  • Ion-selective electrodes (direct potentiometric measurement)

Water of crystallisation changes the molar mass of the solid but not the ion ratio produced upon dissolution. If molar solubility is obtained from mass dissolved, you must use the hydrate’s formula mass to convert mass to moles correctly.

If an ion is basic or acidic (e.g., $CO_3^{2-}$, $S^{2-}$, $F^-$), it can react with water or $H^+$, reducing the free ion concentration used in $K_{sp}$. You may need to distinguish total dissolved species from the equilibrium free ion $[,]$.

Practice Questions

(1–3 marks) Silver chloride has molar solubility s=1.3×105 mol dm3s = 1.3\times10^{-5}\ \text{mol dm}^{-3} at a fixed temperature. Write an expression for KspK_{sp} and determine KspK_{sp} in terms of ss.

  • Writes dissolution: AgCl(s)Ag+(aq)+Cl(aq)AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq) (1)

  • Writes Ksp=[Ag+][Cl]K_{sp}=[Ag^+][Cl^-] (1)

  • Uses [Ag+]=s, [Cl]=s[Ag^+]=s,\ [Cl^-]=s so Ksp=s2=1.7×1010K_{sp}=s^2=1.7\times10^{-10} (1)

(4–6 marks) A saturated solution of CaF2CaF_2 is analysed and found to contain [F]=2.4×104 mol dm3[F^-]=2.4\times10^{-4}\ \text{mol dm}^{-3}. Calculate KspK_{sp} for CaF2CaF_2 at this temperature, stating any stoichiometric relationships you use.

  • Writes dissolution: CaF2(s)Ca2+(aq)+2F(aq)CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq) (1)

  • Correct KspK_{sp} expression: Ksp=[Ca2+][F]2K_{sp}=[Ca^{2+}][F^-]^2 (1)

  • Stoichiometric link: [Ca2+]=12[F][Ca^{2+}]=\frac{1}{2}[F^-] (or equivalent reasoning) (2)

  • Substitutes: Ksp=(122.4×104)(2.4×104)2K_{sp}=\left(\frac{1}{2}\cdot2.4\times10^{-4}\right)(2.4\times10^{-4})^2 (1)

  • Calculates Ksp=6.9×1012K_{sp}=6.9\times10^{-12} (allow sensible rounding) (1)

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