Test: Solubility Equilibria - MCAT MCQ

It should be emphasized that these common polyatomic ions and their charges should be memorized for the MCAT. We may not be accustomed to looking at the Lewis dot structures for these polyatomic ions.

In analyzing the nitrite ion, NO₂ when oxygen has 1 bond, it has a negative charge. When N has 3 and O has 2 bonds, they are uncharged, leaving nitrite ion with an overall charge of -1. It is worth noting that both nitrate (NO₃) and nitrite (NO₂), and have a charge of -1.

In analyzing the phosphite ion, PO₃, since there are 3 single-bonded oxygen, they have a negative charge of -3. Phosphorus is in the same group as nitrogen, and so similarly, when P has 3 bonds, it is uncharged, leaving phosphite with an overall charge of -3. It is worth noting that both phosphate (PO₄) and phosphite (PO₃) have a charge of -3.

In analyzing the sulfite ion, SO₃, we have an element that perpetually expands its octet. Sulfur has 6 valence electrons, so it can form up to 6 bonds. Since 4 are in single bonds and 2 in the lone pair, its formal charge is zero.

Applying the same logic, oxygen has 6 valence electrons. Since 1 is in the single bond and 6 are in the lone pairs, its formal charge is -1, leaving sulfite with an overall charge of -2. It is worth noting that both sulfate (SO₄) and sulfite (SO₃) have a charge of -2.

Lastly, chromate (CrO₄) must be memorized separately with a charge of -2. Therefore, the correct answer is III and IV. For sake of thoroughness, all the oxyhalide anions, perchlorate (ClO₄), chlorate (ClO₃), chlorite (ClO₂), and hypochlorite (ClO), all have a charge of -1.

CASH-N-GIA is a mnemonic for solubility rules. Chlorates, Acetates, Sulfates, Halides (not Fluorides), Nitrates, Group I metals, Ammonium. These compounds are generally soluble.

Lead chloride, PbCl₂ is a halide or chloride, but it is one of the exceptions which include mercury Hg₂²⁺, silver Ag⁺, and lead Pb²⁺. Think HAPpy.

Calcium sulfate, CaSO₄, is a sulfate, but it is one of the exceptions which include the HAPpy group from above, but also CBS for the Group II metals, calcium Ca²⁺, barium Ba²⁺, and strontium Sr²⁺.

The hydroxides are not part of CASHNGIA and should be considered insoluble. The expected exceptions are the Group I metals and ammonium. Aluminum hydroxide, Al(OH)₃, therefore, is considered insoluble.

The chromates are also not part of CASHNGIA, but cesium Cs is a Group I metal, and therefore, caesium chromate, Cs₂CrO₄ is considered soluble.

The correct answer is Cs₂CrO₄.

We are starting with a saturated solution of potassium nitrate rather than pure water. A saturated solution is a solution in which the maximum amount of solute has been dissolved.

That means more of neither the potassium cation or nitrate anion can be dissolved into the solution. Any compounds with either the potassium or nitrate ion will cause precipitate to form.

KOH, Hg₂(NO₃)₂ and NH₄NO₃ will cause a precipitate to form, and only Ag₂SO₄ will not cause precipitate to form. 

If we look at the chart, all the gases (HCl and NH₃) decreases in solubility as the temperature increases. When gases dissolve in a liquid, it is an exothermic process due to the latent heat released when the gas is brought from the gaseous to liquid or aqueous phase.

As for HCl, there is no hydrogen bonding since only F, O, and N can H-bond. Additionally, gas molecules do not H-bond since the particles are far apart.

Most of the solids increases its solubility as the temperature increase because the dissolution of solids is an endothermic process. Heat would be on the reactant side, and by adding more reactant, the equilibrium shifts to the right.

When considering the dissolution of solids, there are 2 possibilities: endothermic and exothermic. When dissolution is exothermic, more will not dissolve when the temperature increases.

For cesium sulfate, the heat released when new bonds are formed between the ions and water molecules or the enthalpy of hydration is very exothermic. It exceeds the energy required to break the lattice, which is equal to the lattice energy.

The correct answer is that the solubility of cesium sulfate decreases since the enthalpy of hydration releases more heat than is needed for the lattice dissociation energy.

Apply Le Châtelier’s principle to the below solubility equilibrium. Depending on the stress that is applied on the system, more may precipitate or dissolve.

By adding NaOH to the solution, the concentration of hydroxide ion will increase. When the concentration of a product increases, the reaction shifts back to the left, and here, more Cu(OH)₂ will precipitate.

By adding copper phosphate to the solution, there will be little effect on the equilibrium since Cu₃(PO₄)₂ is an insoluble solid.

Increasing the pH is the same as increasing the basicity, which can be achieved by adding more hydroxide ion or removing hydronium ion. In either case, more Cu(OH)₂ will not dissolve.

The goal of adding ammonia was not to increase the basicity of the solution, but act as a ligand that binds to a central metal atom to form a coordination complex. The halides and pseudohalides are important anionic ligands whereas ammonia, carbon monoxide, and water are common charge-neutral ligands.

By adding ammonia to the solution, copper forms a complex ion with ammonia to become the copper ammine complex [Cu(NH₃)₄]2+. This will remove Cu²⁺ from the solution, and more Cu(OH)₂ will dissolve.

The correct answer is to add ammonia to the solution to allow more cupric hydroxide to dissolve.

The direct measure of how much solid dissolves is the molar solubility which is expressed in moles solute per liter of solution. The solubility product constant represents the product of the concentrations of the ions present in solution.

The ion product is the product of the concentrations of the ions at any moment in time. When the ion product is greater than the solubility product constant, the solution is considered unsaturated

The molar solubility is always greater than the solubility product constant. Since we are dealing only with sparingly soluble solutes, the constant is expressed with negative exponents. Multiplying two numbers with negative exponents will always yield a smaller number.

Each concentration in the K_sp expression is raised to the stoichiometric coefficient, which happens to the same as the power of the charge of the opposing ion.

This is a graph of the inverse relationship between Ag⁺ and Cl⁻, as expressed by the solubility product constant expression, K_sp = (Ag⁺)(Cl⁻).

All the points that make up the graph represents all the possible concentrations of Ag⁺ and Cl⁻ such that their product equals K_sp

Any points above or to the right of the graph would produce ion products of supersaturated solutions (Q > K_sp). Any points below or to the left of the graph would produce ion products of unsaturated solutions (Q < K_sp).

Point D does have a greater concentration of chloride ions dissolved than point A, but only point D has a K_sp value.

Point E does not have the greatest concentration of ions dissolved, but point C. It is the only point that is the supersaturated area.

Point B does not have a greater concentration of ions dissolved than at point C since point C is a supersaturated solution.

Points B and D are both in equilibrium since they lie on the graph. Point B could represent the case where there is an existing concentration of Ag⁺ in the solution. Point D represents the point where the two ion concentrations are equal.

The temptation is to use the solubility product constants (K_sp) as a direct measure of concentration, but the proper method is to compare the molar solubilities. K_sp values can be used when all the compounds dissociate into the same number of ions, but since the solids here dissociate into a different number of ions, the molar solubilities should be calculated.

Here is a shortcut to determine the rank without much math. Divide the exponent by the number of ions to give you a rough estimate of the magnitude of the negative exponent. For instance, for compound I, divide -22 by 4 to obtain -5.5.

Continue for the rest of the compounds. For compound II, divide -22 by 3 to get -7.3. For compound III, divide -39 by 4 to get -9.75. For compound IV, divide -13 by 3 to get -4.3. If the powers are within one of each other, then the shortcut should not be used but rather the full math.

Remember that these numbers represent negative exponents. Depending on what the question is asking for, the correct order can be IV →  I →  II →  III or III →→right arrow III →  lI →  I →  IV

The question stem is asking for increasing pH, which means increasing basicity or OH⁻ concentration. Compound III has a molar solubility of about 1^0−9.75 while compound IV about 10^−4.3

Since compound III has the lowest molar solubility, the correct order is the following: III →  lI →  I →  IV

Let’s write out the equation for the solubility product constant:

The formation reaction of a complex ion consists of the metal ion and the ligands on the reactant side and the complex ion on the product side:

Combining the two equations together produces the original equation:

When adding or combining equations together, the equilibrium constants must be multiplied. So here is the correct answer:

To calculate the molar solubility of CaF₂ in a solution containing 0.10 M Ca(NO₃)₂, start with the K_sp equation and then replace the right side with actual concentration values:

4.0 x 10⁻¹¹ = (Ca²⁺)(F⁻)²
4.0 x 10⁻¹¹ = (0.1 + y)(2y)²

When the existing concentration in the solution and the K_sp value differ by more than a power of 3, which means 10³ or 1000, the equation can be simplified by approximating (0.1 + y) as 0.1. Ultimately the value of y would be 1000 times less compared to 0.10 M and can be ignored; for instance, 0.1003 or 0.1009 would round to 0.10.

After the approximation, isolate the variable y by dividing both sides by 0.4 and finding the square root of both sides by taking the square root of 1 and dividing the exponent by 2:

To calculate the molar solubility of CaF₂ in a solution containing 0.10 M NaF, start with the equation:

4.0 x 10⁻¹¹ =  (y)(0.1 + 2y)²

Simplify the equation by approximating (0.1 + 2y) to be 0.1 and solve for y by dividing both sides of the equation by 0.01:

4.0 x 10⁻¹¹ =  (y)(0.1)²
y = 4.0 x 10⁻⁹ M

Note the vast difference in solubility due to the common ion effect. 1.0 x 10⁻⁵ M and 4.0 x 10⁻⁹ M are the two molar solubilities, and they differ by a factor of about 2500. So CaF₂ is approximately 2500 times more soluble in a solution containing 0.10 M Ca(NO₃)₂ than in a solution containing 0.10 M NaF.