Solubility - AP Chemistry

Though the solubility of calcium hydroxide, Ca(OH)2, is fairly low, it is a reactant and will not form a precipitate. The solid calcium hydroxide will be added to an aqueous solution of potassium chloride, KCl. During the reaction, the calcium hydroxide will transition to potassium hydroxide (KOH) and calcium chloride (CaCl2), both of which are completely soluble. At the end of the reaction, no precipitate will be observed.

Use the balanced reaction of the dissociation to assign values of X for the solubility of each ion. As CaF2 dissociates, each ion amount will increase by some amount; calcium will increase by X and flouride will increase by 2_X_.

Then plug in the equation to solve Ksp, using the X values assigned for each solute. Use the given value for solubility to find Ksp. Remember that NaCl si not included in the equation, as it is a pure solid.

Standard solubility rules tell us that group I elements and ammonium cations will always result in soluble salts. Lithium, sodium, and potassium are all group I elements, indicating that none of the given answer options will form a precipitate in solution.

Henry's law can be expressed:

Where is the Henry's Law constant, and is the partial pressure of the gas above the solution. We can substitute our givens into this equation:

is composed of two types of ions, and . When it dissociates, it forms:

A fully saturated solution is at equilibrium between the solid and aqueous states, but it is a dynamic equilibrium. As the reacting chemical reacts with aqueous X, more and more of the solid will dissolve into solution.

Lead (II) hydroxide dissociates according to this reaction: . Solubility is equal to the number of moles that will dissociate at equilibrium, and can be found using this reaction and the value of Ksp.

Ksp = \[Pb2+\]\[OH-\]2

If \[Pb2+\] = x, then \[OH-\] = 2x. They exist in a 1:2 ratio.

Ksp = 1.43 * 10-20 = x(2x)2 = 4x3

x = 1.53 x 10-7M

Increasing the temperature of a solution will generally increase its solubility, while decreasing temperature will have the opposite effect. The common ion effect refers to the phenomenon when two compunds in a solution dissociate to produce at least one similar ion, and will make it harder for additional amounts of that ion in the solution to dissociate, decreasing solubility.

First, set up a balanced reaction and enter X for the amount of each solute that is going to be dissolved. As NaCl dissociates, each ion amount will increase by some amount, X.

Then set up the equation for the value of Ksp and enter in the value of X for each solute. Use this equation to solve for X, the solubility (in mol/L). Remember that NaCl si not included in the equation, as it is a pure solid.

Iron (II) has a positive two charge: .

Phosphate has a negative three charge: .

The initial compound must be constructed to cancel these charges. The dissociation is: .

The key to solubility is memorizing the solubility rules. We know that all compounds of alkali metals are soluble. These include potassium, lithium, and sodium, eliminating three of the answers. We also know that all chloride salts are soluble, except those of silver, lead, and mercury. Silver chloride is, thus, not soluble.

All hydroxide compounds are generally insoluble, but alkali metals and ammonium salts will make them soluble. Alkaline earth metals make hydroxides slightly soluble.

When writing the equilibrium expression for an insoluble salt, remember that pure solids and liquids are not included in the expression. Also, the coefficients for the compounds in the balanced reaction become the exponents for the compounds in the equilibrium expression.

Given a generalized chemical reaction, we can determine the equilibrium constant expression.

In our reaction, the reactant is a pure solid and is not included in the equilibrium calculation.

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated and , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

Before we solve for the solubility, remember Le Chatelier's principle. Barium nitrate is a soluble salt that will completely dissolve in solution. This means that there will be a one molar concentration of barium ions before the salt dissolves. Since we have added to the products side of the reaction, we can predict that the equilibrium will be shifted to the left, and the barium hydroxide will be less soluble. This change in equilibrium is referred to as the common ion effect.

We will use an ICE table to determine the solubility of the salt.

I. Before the salt begins to dissolve, there is a one molar concentration of barium ions (from the barium nitrate) with no hydroxide ions in the solution.

C. Since every dissolved molecule of barium hydroxide will yield one barium ion and two hydroxide ions, the unknown increases in each ion can be designated and respectively. Notice how the barium ion increase will be added to the previously added one molar concentration in the solution.

E. Now, we set the solubility expression equal to the solubility product constant.

Since 1 is such a larger number than the value we will get for , we can consider the value in the barium ion expression to be negligible. This allows us to simplify the equation to the following setup.

The solubility of barium hydroxide without the common ion effect is 0.108M. Notice how the solubility for the salt has decreased, as was predicted in the beginning by Le Chatelier's principle.

The solubility product constants help us envision which ions will have the largest concentrations after dissolving to equilibrium. has the largest solubility product constant, meaning it will dissolve the most out of the four options.

Upon the dissolving of , twice the number of fluoride ions will be released as magnesium ions; however, the solubility constant for magnesium fluoride is so much smaller, compared to the solubility constant of , that it will not become the ion with the largest concentration.

Unless paired with an alkali metal, carbonate compounds are generally insoluble. Compounds that contain nitrate or an alkali metal will generally be soluble in water, and hydroxides are soluble when paired with heavier alkaline earth metals (such as calcium).

This question requires an understanding of solubility guidelines and which ion combinations will result in precipitates. Virtually all ionic compounds containing ammonium, sodium, and nitrate will be soluble in water. Most ionic compounds containing chlorine are soluble, with the exceptions of silver, mercury, and lead cations.

If sodium chloride dissolves in this solution, the silver cations and chlorine anions will combine and result in a white, crystalline silver chloride precipitate.

Solubility rules determine which ionic compounds are soluble. All nitrates and group I compounds (those containing alkali metals) are soluble. This means that compounds I and II must be soluble; compound I is a nitrate and compound II contains sodium, an alkali metal.

Compound III, calcium fluoride, is not soluble. Most fluoride compounds are soluble, with the exceptions of: .

This question is asking us to identify a compound that is the least soluble in the solvent water. To be able to answer this question, it's important to understand that, as a general rule of thumb, like dissolves like. In other words, since water is a polar molecule, the compounds that will best dissolve in water are those that are themselves either polar or carry a net charge.

is expected to be soluble because it is polarized. With a lone electron pair on the nitrogen atom and the large electronegativity difference between the nitrogen and each of the three hydrogen atoms, ammonia will easily dissolve within water.

Likewise, is expected to dissolve in water because it can ionize. Once dissolved in water, can dissociate to become and . And since each of these molecules carry a charge and are thus very stabilized within the polar environment that water provides, the dissociation process is extremely favorable.

For much the same reasons, is also soluble in water. It can dissociate into one and two , making it dissolve readily in water.

But what about ? The difference in electronegativity between carbon and chlorine is high, so the compound is expected to be polar, right? Well, not exactly. There is a very important distinction to be made here. Just because a bond within a molecule is polar doesn't mean that the molecule itself is polar. Sure, the bond between carbon and chlorine is indeed polar. But remember, we also need to take the shape of the molecule into account. is a tetrahedral molecule. As a result, each of the chlorine atoms point away from the central carbon atom in the shape of a tetrahedron. Because of this, all of the individual carbon-chlorine dipoles (from each bond) cancels each other out. As a result of this, the net dipole moment on this compound becomes zero, and thus has no polarity. And since the polarity on this compound is zero, it would not be expected to dissolve very well in water.