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11.5: Solubility Rules

Last updated

Mar 10, 2021
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- 11.4: The Dissolution Process
- 11.6: Molecular and Ionic Equations

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To determine if a salt is soluble or to judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds will be insoluble in aqueous solutions. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble, or insoluble). Table $\PageIndex{1}$ lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the (aq) label with it, indicating that it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.

Table $\PageIndex{1}$ : Some Useful Solubility Rules (soluble)
These compounds generally dissolve in water (are soluble):	Exceptions:
All compounds of Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, and NH₄⁺	None
All compounds of NO₃⁻ and C₂H₃O₂⁻	None
Compounds of Cl⁻, Br⁻, I⁻	Ag⁺, Hg₂²⁺, Pb²⁺
Compounds of SO₄²	Hg₂²⁺, Pb²⁺, Sr²⁺, Ba²⁺

Table $\PageIndex{2}$ : Some Useful Solubility Rules (insoluble)
These compounds generally do not dissolve in water (are insoluble):	Exceptions:
Compounds of CO₃²⁻ and PO₄³⁻	Compounds of Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, and NH₄⁺
Compounds of OH⁻	Compounds of Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, NH₄⁺, Sr²⁺, and Ba²⁺

For example, consider the possible double-replacement reaction between Na₂SO₄ and SrCl₂. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble, except for Ag⁺, Hg₂²⁺, and Pb²⁺, which are not being considered here. Therefore, Na₂SO₄ and SrCl₂ are both soluble. The possible double-replacement reaction products are NaCl and SrSO₄. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO₄? Compounds of the sulfate ion are generally soluble, but Sr²⁺ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be:

$\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber$

You would expect to see a visual change corresponding to SrSO₄ precipitating out of solution (Figure $\PageIndex{2}$ ).

Figure $\PageIndex{2}$ : Double-Replacement Reactions. Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution. Source: Photo courtesy of Choij, http://commons.wikimedia.org/wiki/File:Copper_solution.jpg.

Example $\PageIndex{4}$ :

Will a double-replacement reaction occur? If so, identify the products.

Ca(NO₃)₂ + KBr → ?
NaOH + FeCl₂ → ?

Solution

According to the solubility rules, both Ca(NO₃)₂ and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr₂ and KNO₃. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
According to the solubility rules, both NaOH and FeCl₂ are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)₂. NaCl is soluble, but, according to the solubility rules, Fe(OH)₂ is not. Therefore, a reaction would occur, and Fe(OH)₂(s) would precipitate out of solution. The balanced chemical equation is $\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber$

Exercise $\PageIndex{4}$

$\ce{Sr(NO3)2 + KCl → }\nonumber$

Answer: No reaction; all possible products are soluble.

Key Takeaways

Solubility rules are used to predict whether some double-replacement reactions will occur.

Key Takeaways

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