Introduction to Solubility Equilibria (Ksp)溶解平衡与 Ksp

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 7~45 min

Learning goals

  • Write the dissolution equilibrium and the Ksp expression for a slightly soluble salt
  • Calculate molar solubility from Ksp, or Ksp from solubility, accounting for the coefficients of the ions
  • Compare the solubilities of salts of different formula types, and compare the ion product Q with Ksp to predict whether a precipitate forms on mixing

Before you start

  • Solubility rules and precipitation (2.7.3)
  • Solubility (U3.10)
  • Reaction quotient and equilibrium constant (U7.3)

Dissolution equilibrium of a slightly soluble salt

“Insoluble” doesn’t mean none at all dissolves. Put AgCl in water and a very small amount dissolves; once the solution is saturated, dissolving and precipitating happen at the same rate:

AgCl(s)⇌AgX+(aq)+ClX−(aq)\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}

The solid is left out of the expression, so K for this equilibrium is just the product of the ion concentrations, called the solubility product溶度积, Ksp: Ksp = [Ag⁺][Cl⁻]. In general, for MₘXₙ(s) ⇌ mMⁿ⁺(aq) + nXᵐ⁻(aq), Ksp = [Mⁿ⁺]ᵐ[Xᵐ⁻]ⁿ. The Ksp values on this site come from OpenStax Chemistry 2e, Appendix J (25 °C).

Ksp and molar solubility

If s mol of salt dissolves per liter of saturated solution (the molar solubility摩尔溶解度), then [Mⁿ⁺] = m·s and [Xᵐ⁻] = n·s:

Formula typeExamplesKsp and s
MXAgCl, CaCO₃Ksp = s²
MX₂ or M₂XCaF₂, Ag₂CO₃Ksp = (s)(2s)² = 4s³
M₃X or MX₃Ag₃PO₄, Fe(OH)₃Ksp = 27s⁴

For salts of the same formula type, the smaller Ksp means the less soluble salt; for different types you can’t compare Ksp values directly and must calculate s first.

When two solutions are mixed, substituting the ion concentrations after mixing into the expression gives the ion product离子积, Q: if Q > Ksp the solution is supersaturated and a precipitate forms; if Q < Ksp no precipitate forms.

Choose a salt to see each step from Ksp to solubility; then try mixing two solutions:

Try it: solubility product and solubility

Example 1: Solubility from Ksp

At 25 °C, Ksp for PbI₂ is 1.4 × 10⁻⁸. Find the molar solubility of PbI₂ in water, [I⁻] in the saturated solution, and the mass that dissolves per liter (M = 461.0 g/mol).

Step-by-step solution

  1. Write the equilibrium and the Ksp expression.

    PbIX2(s)⇌PbX2+(aq)+2 IX−(aq)Ksp=[PbX2+][IX−]2\begin{gathered} \ce{PbI2(s) <=> Pb^2+(aq) + 2I-(aq)} \\ K_{sp} = [\ce{Pb^2+}][\ce{I-}]^2 \end{gathered}

  2. Express the ion concentrations in terms of s. If s mol of PbI₂ dissolves per liter, it gives s mol of Pb²⁺ and 2s mol of I⁻ (concentrations in M):

    PbI₂(s)Pb²⁺2I⁻
    I—00
    C−s+s+2s
    E—s2s

  3. Substitute into Ksp. When 2s is squared, the 2 is squared too:

    Ksp=s⋅(2s)2=4s34s3=1.4×10−8\begin{aligned} K_{sp} &= s \cdot (2s)^2 = 4s^3 \\ 4s^3 &= 1.4 \times 10^{-8} \end{aligned}

  4. Solve for s. Divide by 4, then take the cube root; the cube root of 10⁻⁹ is exactly 10⁻³:

    s3=3.5×10−9s=3.53×10−3=1.52×10−3≈1.5×10−3 M\begin{aligned} s^3 &= 3.5 \times 10^{-9} \\ s &= \sqrt[3]{3.5} \times 10^{-3} = 1.52 \times 10^{-3} \\ &\approx \boxed{1.5 \times 10^{-3}\ \text{M}} \end{aligned}

  5. Find [I⁻].

    [IX−]=2s=3.0×10−3 M[\ce{I-}] = 2s = \boxed{3.0 \times 10^{-3}\ \text{M}}

  6. Convert to mass per liter. Molar solubility (mol/L) × molar mass (g/mol) = grams dissolved per liter:

    1.52×10−3×461.0=0.70 g/L1.52 \times 10^{-3} \times 461.0 = \boxed{0.70\ \text{g/L}}

Example 2: Ksp from solubility

At 25 °C, the solubility of Ag₂CO₃ is measured as 0.035 g/L (M = 275.75 g/mol). Find Ksp.

Step-by-step solution

  1. Convert the mass solubility to molar solubility. Mass solubility (g/L) ÷ molar mass (g/mol):

    s=0.035275.75=1.27×10−4 Ms = \frac{0.035}{275.75} = 1.27 \times 10^{-4}\ \text{M}

  2. Write the equilibrium and express the ions in terms of s. One Ag₂CO₃ gives 2 Ag⁺ and 1 CO₃²⁻:

    AgX2COX3(s)⇌2 AgX++COX3X2−[AgX+]=2s,[COX3X2−]=s\begin{gathered} \ce{Ag2CO3(s) <=> 2Ag+ + CO3^2-} \\ [\ce{Ag+}] = 2s, \quad [\ce{CO3^2-}] = s \end{gathered}

  3. Write Ksp and substitute s.

    Ksp=[AgX+]2[COX3X2−]=(2s)2⋅s=4s3\begin{aligned} K_{sp} &= [\ce{Ag+}]^2[\ce{CO3^2-}] \\ &= (2s)^2 \cdot s = 4s^3 \end{aligned}

  4. Substitute the value.

    Ksp=4×(1.27×10−4)3=4×2.05×10−12=8.2×10−12\begin{aligned} K_{sp} &= 4 \times (1.27 \times 10^{-4})^3 \\ &= 4 \times 2.05 \times 10^{-12} \\ &= \boxed{8.2 \times 10^{-12}} \end{aligned}

  5. Check. This agrees with the OpenStax table value, 8.1 × 10⁻¹². Two common mistakes: writing [Ag⁺] as s gives Ksp = s³ = 2.0 × 10⁻¹², 4 times too small; forgetting to square the 2 gives 2s³ = 4.1 × 10⁻¹², 2 times too small.

Example 3: Will a precipitate form, and what is left?

0.100 L of 0.200 M K₂C₂O₄ (potassium oxalate) is added to 0.150 L of 0.250 M BaBr₂. Ksp for BaC₂O₄ is 2.3 × 10⁻⁸ (the value given in the problem). Describe what the final mixture looks like, and support your answer with calculations.

Step-by-step solution

  1. Find the possible precipitate. After mixing, the solution contains K⁺, C₂O₄²⁻, Ba²⁺ and Br⁻. All potassium salts are soluble, so KBr does not precipitate; the only possible precipitate is BaC₂O₄:

    BaCX2OX4(s)⇌BaX2++CX2OX4X2−Ksp=[BaX2+][CX2OX4X2−]\begin{gathered} \ce{BaC2O4(s) <=> Ba^2+ + C2O4^2-} \\ K_{sp} = [\ce{Ba^2+}][\ce{C2O4^2-}] \end{gathered}

  2. Concentrations after mixing (don’t forget dilution). The total volume is 0.100 + 0.150 = 0.250 L. First find the moles, then divide by the total volume to get molarity:

    n(CX2OX4X2−)=0.100×0.200=0.0200n(BaX2+)=0.150×0.250=0.0375\begin{aligned} n(\ce{C2O4^2-}) &= 0.100 \times 0.200 \\ &= 0.0200 \\ n(\ce{Ba^2+}) &= 0.150 \times 0.250 \\ &= 0.0375 \end{aligned} [CX2OX4X2−]=0.02000.250=0.0800[BaX2+]=0.03750.250=0.150\begin{aligned} [\ce{C2O4^2-}] &= \frac{0.0200}{0.250} = 0.0800 \\ [\ce{Ba^2+}] &= \frac{0.0375}{0.250} = 0.150 \end{aligned}

    M₁V₁ = M₂V₂ gives the same result: 0.200 × 0.100 = M₂ × 0.250, so M₂ = 0.0800 M.

  3. Calculate Q and compare it with Ksp.

    Q=[BaX2+][CX2OX4X2−]=0.150×0.0800=1.20×10−2\begin{aligned} Q &= [\ce{Ba^2+}][\ce{C2O4^2-}] \\ &= 0.150 \times 0.0800 = 1.20 \times 10^{-2} \end{aligned}

    Q is about 5 × 10⁵ times Ksp (2.3 × 10⁻⁸). Q ≫ Ksp, so the solution is far supersaturated and BaC₂O₄ must precipitate.

  4. Treat the precipitation as complete and find the limiting ion. Ba²⁺ + C₂O₄²⁻ → BaC₂O₄(s) uses the two ions 1 : 1. There is less C₂O₄²⁻ (0.0200 mol) than Ba²⁺ (0.0375 mol), so C₂O₄²⁻ is limiting. A BCA table (U4.5) keeps track of the amounts, in mol:

    Ba²⁺C₂O₄²⁻BaC₂O₄
    Before0.03750.02000
    Change−0.0200−0.0200+0.0200
    After0.0175≈ 00.0200
    m(BaCX2OX4)=0.0200×225.35=4.51 g[BaX2+]=0.01750.250=0.0700 M\begin{aligned} m(\ce{BaC2O4}) &= 0.0200 \times 225.35 \\ &= 4.51\ \text{g} \\ [\ce{Ba^2+}] &= \frac{0.0175}{0.250} \\ &= 0.0700\ \text{M} \end{aligned}

  5. Find how much C₂O₄²⁻ is left. [C₂O₄²⁻] is not really zero: the liquid above the solid is a saturated solution of BaC₂O₄, so a tiny amount of BaC₂O₄ dissolves. Let x M dissolve and set up an ICE table (concentrations in M):

    BaC₂O₄(s)Ba²⁺C₂O₄²⁻
    I—0.07000
    C−x+x+x
    E—0.0700 + xx

    Because x is tiny, 0.0700 + x ≈ 0.0700:

    (0.0700+x) x=2.3×10−80.0700 x≈2.3×10−8x=3.3×10−7 M\begin{aligned} (0.0700 + x)\,x &= 2.3 \times 10^{-8} \\ 0.0700\,x &\approx 2.3 \times 10^{-8} \\ x &= 3.3 \times 10^{-7}\ \text{M} \end{aligned}

    Check: 3.3 × 10⁻⁷ ≪ 0.0700, so the approximation is fine. The excess Ba²⁺ is a common ion that keeps even less BaC₂O₄ in solution (next lesson, U7.12).

  6. List the final ion concentrations (in M). K⁺ and Br⁻ take no part in the reaction; they are only diluted:

    [KX+]=2×0.02000.250=0.160[BrX−]=2×0.03750.250=0.300[BaX2+]=0.0700[CX2OX4X2−]=3.3×10−7\begin{aligned} [\ce{K+}] &= \frac{2 \times 0.0200}{0.250} = 0.160 \\ [\ce{Br-}] &= \frac{2 \times 0.0375}{0.250} = 0.300 \\ [\ce{Ba^2+}] &= 0.0700 \\ [\ce{C2O4^2-}] &= 3.3 \times 10^{-7} \end{aligned}

    Check that charge balances:

    positive=0.160+2×0.0700=0.300negative=0.300+2×3.3×10−7≈0.300 ✓\begin{aligned} \text{positive} &= 0.160 + 2 \times 0.0700 \\ &= 0.300 \\ \text{negative} &= 0.300 + 2 \times 3.3 \times 10^{-7} \\ &\approx 0.300 \ \checkmark \end{aligned}

  7. Describe the appearance. A white precipitate (BaC₂O₄) forms as soon as the two colorless solutions are mixed. Once it settles, about 4.5 g of white solid sits at the bottom under a clear, colorless solution (all the remaining ions are colorless). The oxalate left in solution:

    n=3.3×10−7×0.250=8.2×10−8 mol\begin{aligned} n &= 3.3 \times 10^{-7} \times 0.250 \\ &= 8.2 \times 10^{-8}\ \text{mol} \end{aligned} 8.2×10−80.0200=4×10−6=0.0004%\frac{8.2 \times 10^{-8}}{0.0200} = 4 \times 10^{-6} = 0.0004\%

    The oxalate has been precipitated almost completely, which is called quantitative precipitation.

Practice: at 25 °C in water, which salt has the larger molar solubility? (Ksp in parentheses)
  1. AgCl (1.6 × 10⁻¹⁰) or AgBr (5.0 × 10⁻¹³)
  2. CaF₂ (4.0 × 10⁻¹¹) or AgCl (1.6 × 10⁻¹⁰)
  3. Ag₂CO₃ (8.1 × 10⁻¹²) or CaCO₃ (8.7 × 10⁻⁹)
  4. PbI₂ (1.4 × 10⁻⁸) or PbSO₄ (1.3 × 10⁻⁸)
  5. AgI (1.5 × 10⁻¹⁶) or AgBr (5.0 × 10⁻¹³)
  6. Ag₃PO₄ (1.8 × 10⁻¹⁸) or AgCl (1.6 × 10⁻¹⁰)
  7. Fe(OH)₃ (4 × 10⁻³⁸) or AgI (1.5 × 10⁻¹⁶)
  8. CaCO₃ (8.7 × 10⁻⁹) or Mg(OH)₂ (8.9 × 10⁻¹²)

Common mistakes

Mistake 1: “The smaller Ksp, the lower the solubility, always.” That holds only for salts of the same formula type. For different types, calculate s first and then compare.

Mistake 2: “For CaF₂, Ksp = s × 2s.” [F⁻] is 2s and is squared in the expression: Ksp = s(2s)² = 4s³.

Mistake 3: “Adding more solid to a saturated solution raises the ion concentrations.” The solid doesn’t affect the equilibrium; Ksp sets the ion concentrations in a saturated solution, and the extra solid just sits on the bottom.

Quick check

What is the Ksp expression for Ca₃(PO₄)₂?

At 25 °C, Ksp for AgBr is 5.0 × 10⁻¹³ (M = 187.8 g/mol). What is the greatest mass of AgBr that can dissolve in 2.0 L of water?

The molar solubility of CaF₂ is measured as 2.2 × 10⁻⁴ M. The Ksp calculated from this is about:

Equal volumes of 2.0 × 10⁻⁴ M AgNO₃ and 2.0 × 10⁻⁴ M NaCl are mixed (Ksp of AgCl = 1.6 × 10⁻¹⁰). Does a precipitate form?

For which salt is the molar solubility calculated with s = (Ksp/27)¼?

More solid AgCl is added to a saturated AgCl solution (with AgCl solid on the bottom), stirred, and left to settle. [Ag⁺] in the solution:

AP-style free response

AP-style free response

At 25 °C, Ksp for Mg(OH)₂ is 8.9 × 10⁻¹² and Ksp for CaCO₃ is 8.7 × 10⁻⁹ (OpenStax).

(a) Write the dissolution equilibrium and the Ksp expression for Mg(OH)₂.

(b) Calculate the molar solubility of Mg(OH)₂ in water and [OH⁻] in the saturated solution.

(c) Ksp for CaCO₃ is about 1000 times larger than for Mg(OH)₂. Which has the larger molar solubility in water? Justify your answer with calculations.

(d) 50.0 mL of 2.0 × 10⁻³ M MgCl₂ is mixed with 50.0 mL of 2.0 × 10⁻³ M NaOH. Does Mg(OH)₂ precipitate?

(e) Explain why comparing the Ksp values alone is not enough in (c).

Model answer and scoring

(a) Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq), Ksp = [Mg²⁺][OH⁻]².

(b) Ksp = 4s³, so s = ³√(8.9 × 10⁻¹² ÷ 4) = 1.3 × 10⁻⁴ M; [OH⁻] = 2s = 2.6 × 10⁻⁴ M.

(c) For CaCO₃, s = √(8.7 × 10⁻⁹) = 9.3 × 10⁻⁵ M, smaller than 1.3 × 10⁻⁴ M for Mg(OH)₂. Mg(OH)₂ has the larger molar solubility.

(d) After mixing, [Mg²⁺] = [OH⁻] = 1.0 × 10⁻³ M, so Q = 1.0 × 10⁻³ × (1.0 × 10⁻³)² = 1.0 × 10⁻⁹ > 8.9 × 10⁻¹²: a precipitate forms.

(e) The two salts are different formula types: Ksp = s² for CaCO₃ and Ksp = 4s³ for Mg(OH)₂. Because Ksp is related to s differently, s must be calculated before comparing.

Scoring: (a) equilibrium and expression; (b) 4s³ and 2s; (c) calculating and comparing both values of s; (d) halved concentrations after mixing, and comparing Q with Ksp; (e) different formula types.

Summary

  • For a slightly soluble salt MₘXₙ, Ksp = [Mⁿ⁺]ᵐ[Xᵐ⁻]ⁿ, with the solid left out.
  • Molar solubility s: Ksp = s² for MX, 4s³ for MX₂, and 27s⁴ for M₃X. For different types, compare s, not just Ksp.
  • If the ion product Q > Ksp, a precipitate forms; if Q < Ksp, it doesn’t.

AP tip: Include the coefficient when you write an ion concentration ([F⁻] = 2s) and square the whole 2s when you substitute. For mixing problems, find the concentrations after mixing (the volume has changed) before calculating Q.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.11 · Introduction to Solubility Equilibria (Ksp) · 5 questions

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