Kaplan MCAT General Chemistry Review

Solubility product constants, like all other equilibrium constants (K_eq, K_a, and K_b) are temperature-dependent. When the solution consists of a gas dissolved into a liquid, the value of the equilibrium constant, and hence the “position” of equilibrium (saturation), will also depend on pressure. Generally speaking, the solubility product constant increases with increasing temperature for nongas solutes and decreases for gas solutes. Higher pressures favor dissolution of gas solutes and therefore the K_sp will be larger for gases at higher pressures than at lower ones. This last point is especially relevant to deep-sea divers. Because gases become more soluble in solution as pressure increases, a diver who has spent time at significant depths under water will have more nitrogen gas dissolved in her blood. (Nitrogen gas is the main inert gas in the air we breathe.) If she rises to the surface too quickly, the abrupt decompression will lead to an abrupt decrease in gas solubility in the plasma, resulting in the formation of nitrogen gas bubbles in her bloodstream. The gas bubbles can get lodged in the small vasculature of the peripheral tissue, mostly around the large joints of the body, causing pain and tissue damage (hence the name of the condition is the “bends”). The condition is painful and dangerous, and can be fatal if not properly prevented or treated.

MCAT Expertise

On the MCAT, if you remember that K_sp is just a specialized form of K_eq, then you can simplify a lot of problems by using the same concepts that you do for all equilibria, including Le Châtelier’s principle.

As solute dissolves into solvent the system approaches saturation, at which point no more solute can be dissolved and any excess will precipitate to the bottom of the container. You may not know whether the solution has reached saturation, and so to determine “where” the system is with respect to the equilibrium position, you will calculate a value called the ion product (I.P.), which is analogous to the reaction quotient Q for chemical reactions. The ion product equation has the same form as the equation for the solubility product constant. The difference is that the concentrations that you use are the concentrations of the ionic constituents at that given moment in time.

I.P. = [Aⁿ⁺]^m[B^m-]ⁿ

Key Concept

If the solution is supersaturated, Q_sp > K_sp, precipitation will occur. If the solution is unsaturated, Q_sp < K_sp, the solute will continue to dissolve. If the solution is saturated, Q_sp = K_sp, then the solution is at equilibrium.

where the concentrations are not necessarily equilibrium (saturation) concentrations. As with the reaction quotient Q, the utility of the I.P. lies in comparing its value to that attained at equilibrium, in this case, the known K_sp. Each salt has its own distinct K_sp at a given temperature and pressure. If, at a given set of temperature and pressure conditions, a salt’s I.P. is less than the salt’s K_sp, then the solution is not yet at equilibrium and we say that it is unsaturated. For unsaturated solutions, dissolution is thermodynamically favored over precipitation. If the I.P is greater than the K_sp, then the solution is beyond equilibrium and we say that it is supersaturated. It’s possible to create a supersaturated solution by dissolving solute into a hot solvent and then slowly cooling the solution. A supersaturated solution is thermodynamically unstable, and any disturbance to the solution, like the addition of more solid solute or other solid particles or further cooling of the solution, will cause spontaneous precipitation of the excess dissolved solute. If the calculated I.P. is equal to the known K_sp, then the solution is at equilibrium, the rates of dissolution and precipitation are equal, and the concentration of solute is at the maximum (saturation) value.

Example: The molar solubility of Fe(OH)₃ in an aqueous solution was determined to be 4.5 × 10^-10 mol/L. What is the value of the K_sp for Fe(OH)₃?

Key Concept

Every slightly soluble salt of general formula MX₃ will have K_sp = 27x⁴, where x is the molar solubility.

Solution: The molar solubility (the solubility of the compound in mol/L) is given as 4.5 × 10^-10 M. The equilibrium concentration of each ion can be determined from the molar solubility and the balanced dissociation reaction of Fe(OH)₃. The dissociation reaction is:

Example: What are the concentrations of each of the ions in a saturated solution of PbBr₂, given that the K_sp of PbBr₂ is 2.1 × 10^-6? If 5 g of PbBr₂ are dissolved in water to make 1 L of solution at 25°C, would the solution be saturated, unsaturated, or supersaturated?

Key Concept

Every slightly soluble salt of general formula MX₂ will have K_sp = 4x³, where x is the molar solubility.

Solution: The first step is to write out the dissociation reaction:

PbBr₂(s) Pb²⁺(aq) + 2Br^-(aq)

K_sp = [Pb²⁺][Br^-]²