Learning Objective

The solubility product constant for barium sulfate is Kp = 1.1 x 10'. Beaker 1 Beaker 2 Beaker 3 Beaker 4 Water 0.100 M 0.100 M 0.200 M NaCl Na SO Ba (NO3)2 mmon Ion Effect on Solubility 13 W US22762ULUTUMI/O2UIUNULULITUUUULUUMUULUUUUUUU MICHIELO WHICH precip ocas depende UIT TE SOUL PROCLC COTISAT IE precipitate. ™POGIL Activities for AP. Chemistry Equilibrium Systems Reaction Quotient.181. Common Ion Effect on Solubility When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. The role that the common ion effect plays in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium. The common ion effect also plays a role in the regulation of buffers. Exercise 4 Solubility and Common Ions Calculate the solubility of solid CaF2 (Ksp = 4.0 × 10 −11) in a 0.025 M NaF solution. = 6.4 × 10−8 mol/L Ksp AND THE REACTION QUOTIENT, Q ¾ With some knowledge of the reaction quotient, we can decide whether a ppt.

Calculate the molar solubility of a compound in solution containing a common ion.

Key Points

Adding a common ion decreases the solubility of a solute.
The common-ion effect can be used to separate compounds or remove impurities from a mixture.
Different common ions have different effects on the solubility of a solute based on the stoichiometry of the balanced equation.

Terms

precipitateA solid that exits the liquid phase of a solution.
precipitateTo come out of a liquid solution into solid form.
limestoneAn abundant rock of marine and fresh-water sediments; primarily composed of calcite (CaCO₃); it occurs in a variety of forms, both crystalline and amorphous

Adding a Common Ion

If you have a solution and solute in equilibrium, adding a common ion (an ion that is common with the dissolving solid) decreases the solubility of the solute. This is because Le Chatelier’s principle states the reaction will shift toward the left (toward the reactants) to relieve the stress of the excess product. When equilibrium is shifted toward the reactants, the solute precipitates.

Scientists take advantage of this property when purifying water. In areas where water sources are high in chalk or limestone, drinking water contains excess calcium carbonate CaCO₃. In the water treatment process, sodium carbonate salt is added to precipitate the calcium carbonate. The very pure and finely divided precipitate of calcium carbonate that is generated is used in the manufacture of toothpaste.

Example:

What is the solubility at 25°C of calcium fluoride (CaF₂): (a) in pure water; (b) in 0.10 M calcium chloride (CaCl₂); and (c) in 0.10 M sodium fluoride (NaF)?

Solution:

[latex]CaF_2 leftrightarrow Ca^{2+} + 2F^-[/latex]

(a) If the solubility in pure water is s, then

[Ca²⁺] = s and [F^–] = 2s

Common ion effect on solubility pogil answers pdf

[latex]K_{sp} = {[Ca^{2+}]}{[F^-]}^2[/latex]

K_sp = s(2s)² = 4s³ = 3.9 x 10^-11

s = 2.1 x 10^-4 M

(b) Here the calcium ion concentration is the sum of the concentrations of calcium ions from the 0.10 M calcium chloride and from the calcium fluoride whose solubility we are seeking:

[Ca²⁺] = 0.10 + s

[F^–] = 2s

K_sp = (0.10 + s)(2s)² = 3.9 x 10^-11

Can we simplify this equation? With such a small solubility product for CaF₂, you can predict its solubility << 0.10 moles per liter. If our prediction is valid, we can simplify the solubility-product equation:

K_sp = 0.10 (2s)² = 0.40s² = 3.9 x 10^-11

s² = [latex]frac{3.90 times 10^{-11}}{0.40}[/latex] = 9.75 x 10^-11

s = 9.9 x 10^-6 M

If we go back and compare, only 4.7 percent as much CaF₂ will dissolve in 0.10 M CaCl₂ as in pure water:

[latex]frac{(9.9 times 10^{-6})}{2.1 times 10^{-4}}[/latex] x 100 = 4.7%

Therefore, the approximation that s is small compared to 0.10 M was reasonable.

(c) In 0.10 M NaF:

[Ca²⁺] = s

[F^–] = 0.10 + 2s

since fluoride ions are in NaF as well as in CaF₂.

The solubility-product equation is:

K_sp = s(2s + 0.10)² = 3.9 x 10^-11

Again, the equation can be simplified. The 2s term is << 0.10 moles per liter, and therefore:

K_sp = s(0.10)² = 3.9 x 10^-11

s = 3.9 x 10^-9 M

This approximation is also valid, since only 0.0019 percent as much CaF₂ will dissolve in 0.10 M NaF as in pure water. Fluoride is more effective than calcium as a common ion because it has a second-power effect on the solubility equilibrium.

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“Chemical Principles/Solution Equilibria: Acids and Bases.”

Common Ion Effect On Solubility Pogil Answers

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