MCAT Physical - Solubility and Solution Equilibrium

Which of the following will not increase the solubility of a solution?

Increasing the temperature of a gas solute in a liquid solvent

Increasing the temperature of a solid solute in a liquid solvent

Increasing the pressure of a gas solute in a liquid solvent

Decreasing the temperature of a gas solute in a liquid solvent

All of these will increase solubility

Explanation

Increasing the temperature of a gas solute in a liquid solvent will decrease solubility. While dissolved, the gas is in equilibrium with the liquid. Adding heat will push the equilibrium in favor of the gas, causing it to precipitate from solution in the form of bubbles.

Increasing the pressure of a gas solute in a liquid solvent, decreasing the temperature of a gas solute in a liquid solvent, and increasing the temperature of a solid solute in a liquid solvent will all increase the solubility.

The solubility product constant of is . What is the solubility of this compound?

Explanation

Solubility is the ability of a solute to dissolve into a solvent. It is calculated by creating an equilibrium equation and solving for the concentration of dissolved ions.

For this question, we will need to set up the equilibrium constant equation for calcium hydroxide.

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-;\ K_{sp}=[Ca^{2+}][OH^-]^2$

Note that the solid compound is not included in the equilibrium expression. Now we can work on finding the solubility. For each mole of calcium hydroxide dissolved, one mole of calcium ions and two moles of hydroxide ions are released. Mathematically, we can equate this ratio to the ion concentrations in the equilibrium calculation.

$Ca(OH_2):\ K_{sp}=5.02*10^{-5}=(x)(2x)^2$

In this calculation, $\small x$ is the solubility. Given the solubility constant, we can solve for $\small x$ .

$5.02*10^{-5}=4x^3\rightarrow x_{Ca(OH)_2}=0.023M$

Which substance is more soluble in hydrochloric acid than sodium hydroxide?

$Pb(OH)_{2}$

$CaF_{2}$

Explanation

Though we are asking about solubility in a strong acid and in a strong base, this is not an acid-base chemistry question.

Instead, we are dealing with solubility, which is directly affected by the solubility product constant. The solubility product constant for each answer choice is given below.

$K_{spHI}=[H^+][I^-]$

$K_{spCaF_2}=[Ca^{2+}][F^-]^2$

$K_{spAgCl}=[Ag^+][Cl^-]$

$K_{spPb(OH)_2}=[Pb^{2+}][OH^-]^2$

Hydrochloric acid will contribute to hydrogen and chloride ion concentrations, while sodium hydroxide will contribute to sodium and hydroxide ion concentrations. The solubility product constant for hydroiodic acid incorporates the hydrogen ions, while the product constant for lead (II) hydroxide incorporates the hydroxide ions.

Increasing the ion concentration results in the common ion effect. Compounds are less soluble in solutions already containing ions that contribute to their solubility product constant. We can thus conclude that lead (II) hydroxide will be less soluble in sodium hydroxide because the sodium hydroxide contributes a common ion.

Which of the following is present in a solution of sodium cyanide?

$\small Na^+$ and $\small CN^-$ ions, but almost no $\small NaCN$ molecules

$\small Na^+$ ions and $\small NaCN$ molecules, but no $\small CN^-$ ions

$\small CN^-$ ions and $\small NaCN$ molecules, but no $\small Na^+$ ions

$\small Na^+$ ions, $\small CN^-$ ions, and $\small NaCN$ molecules

$\small NaCN$ molecules, but almost no $\small Na^+$ or $\small CN^-$ ions

Explanation

Salts are a type of strong electrolyte, as are strong acids and strong bases. Strong electrolytes break up completely into ions (or "ionize") when dissolved in water. A solution of sodium cyanide would contain almost entirely sodium and cyanide ions, and no salt molecules.

$NaCN(s)\rightleftharpoons Na^+(aq)+CN^-(aq)$

Which of the following compounds would make a weak electrolyte in solution?

Explanation

A strong electrolyte is a compound that will conduct electricity while in solution. This is because it creates many charged ions in the solution. Because we want a weak electrolyte, the answer will be the compound that does not form many ions when in solution. NH3 is a weak base which does not dissociate completely. As a result, it creates few ions in solution, making it a weak electrolyte.

The other answer choices represent either ionic compounds or strong acids, and will dissociate completely in solution.

Scandium hydroxide, , has a $K_{sp}$ value of $8.0*10^{-31}$ . What is its solubility?

$1.3*10^{-8}$

$5.2*10^{-16}$

$3.1*10^{-11}$

$5.3*10^{-15}$

$3.0*10^{-16}$

Explanation

Scandium hydroxide dissociates according to the following reaction.

$Sc(OH)_{3}(s) \rightarrow Sc^{3+}(aq) +3(OH)^{-}(aq)$

The $K_{sp}$ for this dissociation is written as follows. Note that the solid is left out of the equation.

$Ksp = [Sc^{3+}][OH^{-}]^{3}$

If is the amount of $Sc^{3+}$ that forms when dissociates, then is the amount of formed, based on stoichiometry.

$8.0 * 10^{-31} = [x][3x]^{3} = 27x^{4}$

Solving this equation gives us the value of .

$2.96*10^{-32}=x^4$

$x = 1.3 * 10^{-8}$

This value is the solubility of scandium hydroxide.

Barium fluoride dissolves in solution according to the following equation.

$BaF_{2}(s) = Ba^{2+}(aq) + 2F^{-}(aq)$

$K_{sp} = 2.4 * 10^{-5}$

Write the solubility product constant (Ksp) for the reaction.

$K_{sp} = [Ba^{2+}][F^{-}]^{2}$

$K_{sp} = \frac{[Ba^{2+}]2[F^{-}]}{[BaF_{2}]}$

$K_{sp} = \frac{[Ba^{2+}][F^{-}]^{2}}{[BaF_{2}]}$

$K_{sp} = [Ba^{2+}][F^{-}]$

Explanation

When writing the solubility product for a reaction, it is important to remember that solid and liquid compounds are not included in the expression. As a result, barium fluoride would not be included, because it is a solid. Just like the law of mass action, coefficients are the exponents in the expression.

Since only the products are used in the expression (the reactant is a pure solid), it is written as $K_{sp} = [Ba^{2+}][F^{-}]^{2}$ .

Which of the following is true of a saturated solution?

The solution is at equilibrium

Salt will spontaneously precipitate out of the solution

The solute will continue to dissolve.

The ionization product is greater than the Ksp

Explanation

A saturated solution is a solution that is at equilibrium. The ionization product is equal to the Ksp in a saturated solution, and the rate of the forward reaction is equal to the rate of the reverse reaction. In an unsaturated solution, the solute continues to dissolve until equilibrium is reached. In a supersaturated solution, salt will "crash out," or precipitate, from solution.

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Given the polarity of O-H bonds in water described in the passage, what is the Ksp expression for the following reaction, assuming it takes place in an aqueous environment?

$CaSO_{4} (s) \rightarrow Ca^{2+} (aq) + (SO_{4}})^{2-} (aq)$

$[Ca^{2+}][(SO_{4})^{2-}}]$

$\frac{[Ca^{2+}][SO_4^{2-}]}{[CaSO_4]}$

$\frac{[CaSO_4]}{[Ca^{2+}][SO_4^{2-}]}$

$\frac{[Ca^{2+}]}{[CaSO_4]}$

$\frac{[Ca^{2+}][SO_4^{2-}]}{[CaSO_4][H_2O]}$

Explanation

The Ksp is a standard equilibrium constant for the dissolution reaction of a solute. As such, it follows the standard formula for equilibrium constants, and excludes pure liquids and solids (like water and calcium sulfate!).

$K_{eq}=K_{sp}=\frac{[Products]}{[Reactants]}$

Hypothetically, if calcium hydroxide, lithium hydroxide, and potassium hydroxide all had a $\small K_{sp}$ value of $\small 4.0*10^{-6}$ , which compound would have the greatest solubility?

$LiOH\ \text{and}\ KOH$

The solubilities of all three compounds are equal

Explanation

Solubility is the ability of a solute to dissolve into a solvent. It is calculated by creating an equilibrium equation and solving for the concentration of dissolved ions.

For this question, we will need to set up equilibrium constant equations for each compound.

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-;\ K_{sp}=[Ca^{2+}][OH^-]^2$

$LiOH\rightleftharpoons Li^++OH^-;\ K_{sp}=[Li^+][OH^-]$

$KOH\rightleftharpoons K^++OH^-;\ K_{sp}=[K^+][OH^-]$

Note that the solid compounds are not included in the equilibrium expressions. Now we can work on finding the solubilities. For each mole of calcium hydroxide dissolved, one mole of calcium ions and two moles of hydroxide ions are released. For each mole of lithium hydroxide dissolved, one mole of lithium ions and one mole of hydroxide ions are released. For each mole of potassium hydroxide dissolved, one mole of potassium ions and one mole of hydroxide ions are released. Mathematically, we can equate these ratios to the ion concentrations in the equilibrium calculations.

$Ca(OH_2):\ K_{sp}=4.0*10^{-6}=(x)(2x)^2$