# Enduring Understanding 6.C.3: Solubility

- The dissolution of a substance in a solvent is a reversible process, and therefore has an equilibrium constant, K
_{sp}associated with it. **Solubility**can be understood in terms of chemical equilibrium.- This can be used to determine if a precipitate forms when two solutions are mixed, or what the concentration of a species in solution will be.
- Sodium, potassium, ammonium, and nitrate salts are almost always highly soluble in water.
- Sample Problem 1: If 100 mL of a 0.010M CaCl
_{2}solution is mixed with 400 mL of a 0.020M Na_{3}PO_{4}solution, (final volume 500 mL), will a precipitate form? The K_{sp}of Ca_{3}(PO_{4})_{2}is 1.0 x 10^{-26} - The resulting ionic equation is:
- So the K
_{sp}will be given by: - The concentration of Ca
^{2+}will be: (0.010M) x (100 mL/500 mL)= 0.0020M - The concentration of PO
_{4}^{3-}will be: (0.020M) x (400 mL/500 mL)= 0.016M - The solubility product Q
_{sp}will be: - Since Q
_{sp}>> K_{sp}, a precipitate of Ca_{3}(PO_{4})_{2(s)}will form. - The solubility of a poorly soluble salt decreases in the presence of a common ion.
- For example, in the system MX
_{(s)}⇆ M^{+}_{(aq)}+ X^{-}_{(aq)}, if extra X^{-}was added, for example as NaX, the solubility of MX_{(s)}would be lower. - Sample Problem 2: The molar solubility of of Ca
_{3}(PO_{4})_{2(s)}in water is 2.5 x 10^{-6}mol/L. What is the molar solubility of Ca_{3}(PO_{4})_{2(s)}in an 0.1M solution of sodium phosphate?
Ca - x moles of Ca
_{3}(PO_{4})_{2(s)}dissolve to form 3x moles of Ca^{2+}and 2x moles of PO_{4}^{3-} - In pure water, we could substitute x this into the K
_{sp}to give:
[Ca - But, this is an 0.1M sodium phosphate solution, so the equation becomes:
- Ca
_{3}(PO_{4})_{2(s)}has a very low solubility. If we assume x << 0.1, this reduces to: - So the molar solubility of Ca
_{3}(PO_{4})_{2(s)}in 0.1M sodium phosphate is 3.3 x 10^{-9}mol/L, which about 1/1000 of the solubility in water.

_{(s)}⇆ M

^{+}

_{(aq)}+ X

^{-}

_{(aq)}K

_{sp}= [M

^{+}] [X

^{-}]

_{2(aq)}+ 2 Na

_{3}PO

_{4(aq)}⇆ Ca

_{3}(PO

_{4})

_{2(s)}+ 6 NaCl

_{(aq)}

_{3}(PO

_{4})

_{2(s)}⇆ 3 Ca

^{2+}

_{(aq)}+ 2 PO

_{4}

^{3-}

_{(aq)}

_{sp}= [Ca

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= 1.0 x 10

^{-26}

_{sp}= [Ca

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= 0.0020

^{3}x 0.016

^{2}= 2.0 x 10

^{-12}

_{3}(PO

_{4})

_{2(s)}⇆ 3 Ca

^{2+}

_{(aq)}+ 2 PO

_{4}

^{3-}

_{(aq)}K

_{sp}= [Ca

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= 1.0 x 10

^{-26}

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= (3x)

^{3}(2x)

^{2}= 1.0 x 10

^{-26}

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= (3x)

^{3}(2x + 0.1)

^{2}= 1.0 x 10

^{-26}

^{2+}]

^{3}[PO

_{4}

^{3-}]

^{2}= (3x)

^{3}(0.1)

^{2}= 1.0 x 10

^{-26}

27 x

^{3}(0.01) = 1.0 x 10

^{-26}

x

^{3}= 3.7 x 10

^{-26}

x = 3.3 x 10

^{-9}

Related Links:Chemistry Chemistry Quizzes AP Chemistry Notes Buffers |

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