AP Chemistry · Lesson 3

Solubility equilibria and Ksp

Know Know it

A slightly soluble ionic solid in water sets up an equilibrium with its ions. For MaXb(s)⇌aMm+(aq)+bXn−(aq)\text{M}_a\text{X}_b(s) \rightleftharpoons a\text{M}^{m+}(aq) + b\text{X}^{n-}(aq), the solubility product is Ksp=[Mm+]a[Xn−]bK_{sp} = [\text{M}^{m+}]^a[\text{X}^{n-}]^b. The solid is left out.

The molar solubility s is the moles of solid that dissolve per liter of saturated solution. Use the ion ratio to link s and KspK_{sp}: for a 1:1 salt MX, Ksp=s2K_{sp} = s^2; for MX2\text{MX}_2, the ions are s and 2s, so Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3.

Comparing KspK_{sp} values tells you which salt is more soluble only when the salts have the same ion ratio. Otherwise, compute s for each.

To predict precipitation, compute the ion product Q from the concentrations after mixing. If Q>KspQ > K_{sp}, a solid forms; if Q<KspQ < K_{sp}, no solid forms; if Q=KspQ = K_{sp}, the solution is just saturated.

Worked example Easy

The molar solubility of a salt MX in water is 1.3×10−41.3 \times 10^{-4} M at 25 °C. What is KspK_{sp} for MX at 25 °C?

  1. 2.6×10−42.6 \times 10^{-4}
  2. 1.7×10−81.7 \times 10^{-8}Answer
  3. 8.8×10−128.8 \times 10^{-12}
  4. 1.3×10−41.3 \times 10^{-4}

How to solve it

  1. MX(s) dissolves into one M+\text{M}^+ and one X−\text{X}^-, so each ion is at s.
  2. Ksp=s⋅s=(1.3×10−4)2=1.7×10−8K_{sp} = s \cdot s = (1.3 \times 10^{-4})^2 = 1.7 \times 10^{-8} (two significant figures).

Why each choice is right or wrong

  • A. Incorrect. 2.6×10−42.6 \times 10^{-4} is 2s, the sum of the two ion concentrations; KspK_{sp} is their product.
  • B. Correct. Each ion is at 1.3×10−41.3 \times 10^{-4} M, so Ksp=(1.3×10−4)2=1.7×10−8K_{sp} = (1.3 \times 10^{-4})^2 = 1.7 \times 10^{-8}.
  • C. Incorrect. 8.8×10−128.8 \times 10^{-12} is 4s34s^3, the relation for a salt of type MX2\text{MX}_2, not for a 1:1 salt.
  • D. Incorrect. 1.3×10−41.3 \times 10^{-4} is just the molar solubility itself; KspK_{sp} is the product of the ion concentrations.

Trap A bigger Ksp is not always more soluble

Comparing KspK_{sp} values works only for salts with the same ion ratio. A salt of type MX2\text{MX}_2 can have a smaller KspK_{sp} than a salt of type MX and still dissolve more, because Ksp=4s3K_{sp} = 4s^3 for it and s2s^2 for the other.

Explain Explain it

Task verbs: calculate, compare

The common-ion effect is Le Chatelier’s principle applied to a solid. A salt dissolves less in a solution that already contains one of its ions. For a 1:1 salt in a solution with a common ion at concentration c (much larger than s), s≈Ksp/cs \approx K_{sp}/c.

If the anion is a base (such as OH−\text{OH}^-, F−\text{F}^- or CO32−\text{CO}_3^{2-}), adding acid removes that anion, so Q falls below KspK_{sp} and more solid dissolves. Salts like these are more soluble at low pH.

Worked example Medium

A salt MX2\text{MX}_2 has Ksp=3.2×10−11K_{sp} = 3.2 \times 10^{-11} at 25 °C. What is its molar solubility in pure water at 25 °C?

  1. 5.7×10−65.7 \times 10^{-6} M
  2. 3.2×10−43.2 \times 10^{-4} M
  3. 4.0×10−44.0 \times 10^{-4} M
  4. 2.0×10−42.0 \times 10^{-4} MAnswer

How to solve it

  1. Ions: [M2+]=s[\text{M}^{2+}] = s and [X−]=2s[\text{X}^-] = 2s.
  2. Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3.
  3. s=(3.2×10−11)/43=8.0×10−123=2.0×10−4s = \sqrt[3]{(3.2 \times 10^{-11})/4} = \sqrt[3]{8.0 \times 10^{-12}} = 2.0 \times 10^{-4} M.

Why each choice is right or wrong

  • A. Incorrect. 5.7×10−65.7 \times 10^{-6} M is Ksp\sqrt{K_{sp}}, which treats the salt as 1:1 even though it releases two X−\text{X}^- ions.
  • B. Incorrect. 3.2×10−43.2 \times 10^{-4} M is Ksp3\sqrt[3]{K_{sp}}, which leaves out the factor 4 that comes from (2s)2(2s)^2.
  • C. Incorrect. 4.0×10−44.0 \times 10^{-4} M is [X−]=2s[\text{X}^-] = 2s, the anion concentration, not the molar solubility of the salt.
  • D. Correct. With Ksp=4s3K_{sp} = 4s^3, s=3.2×10−11/43=2.0×10−4s = \sqrt[3]{3.2 \times 10^{-11}/4} = 2.0 \times 10^{-4} M.

Compare Your turn

Two slightly soluble salts are tested at 25 °C. Salt AB (one A+\text{A}^+ and one B−\text{B}^-) has Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}. Salt A2C\text{A}_2\text{C} (two A+\text{A}^+ and one C2−\text{C}^{2-}) has Ksp=1.1×10−12K_{sp} = 1.1 \times 10^{-12}. Calculate the molar solubility of each salt in pure water, then say which salt has the greater molar solubility.

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Compare with a model answer

A full answer does this

  • Finds the molar solubility of AB, about 1.3×10−51.3 \times 10^{-5} M
  • Finds the molar solubility of A2C\text{A}_2\text{C}, about 6.5×10−56.5 \times 10^{-5} M
  • Says that A2C\text{A}_2\text{C} has the greater molar solubility

Model answer

For AB, Ksp=s2K_{sp} = s^2, so s=1.8×10−10=1.3×10−5s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} M. For A2C\text{A}_2\text{C}, the ions are 2s and s, so Ksp=4s3K_{sp} = 4s^3 and s=1.1×10−12/43=6.5×10−5s = \sqrt[3]{1.1 \times 10^{-12}/4} = 6.5 \times 10^{-5} M. So A2C\text{A}_2\text{C} has the greater molar solubility, even though its KspK_{sp} is smaller.

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