Question

How can you quickly determine the saturation solubility of a sparingly soluble \(1: 1\) salt at \(25^{\circ} \mathrm{C}\) ?

Short answer

To quickly determine the saturation solubility of a sparingly soluble 1:1 salt at 25°C, find the solubility product constant (K\(_{sp}\)) for the salt and write the dissolution equilibrium equation as AB(s) \(\rightleftharpoons\) A\(^{+}\)(aq) + B\(^{-}\)(aq). Then, write the K\(_{sp}\) expression as [A\(^{+}\)][B\(^{-}\)]. If the saturation solubility is s, the equation becomes K\(_{sp}\) = s\(^{2}\). Solve for s by taking the square root of K\(_{sp}\): s = \(\sqrt{K_{sp}}\). This will give the saturation solubility of the 1:1 salt at 25°C.

Step-by-step solution

Identify the solubility product constant (K\(_{sp}\))

To quickly determine the saturation solubility of a sparingly soluble 1:1 salt at 25°C, we need to know the solubility product constant (K\(_{sp}\)) for that salt. This value can be found in a table of K\(_{sp}\) values in a textbook or online. The K\(_{sp}\) value will allow us to determine the concentration of ions in the solution at equilibrium.

Step 2:Write the dissolution equilibrium equation of salt

Write the balanced chemical equation for the dissolution of the salt in water. For a generic sparingly soluble 1:1 salt AB, the equation looks like this: AB(s) \(\rightleftharpoons\) A\(^{+}\)(aq) + B\(^{-}\)(aq)

Write the expression for the solubility product constant (K\(_{sp}\))

Now, we can write the expression for the solubility product constant (K\(_{sp}\)) based on the dissolution equilibrium equation. For the generic equation: K\(_{sp}\) = [A\(^{+}\)][B\(^{-}\)] Here, [A\(^{+}\)] and [B\(^{-}\)] are the equilibrium concentrations of ions A\(^{+}\) and B\(^{-}\) in the solution.

Calculate the saturation solubility using K\(_{sp}\)

Let the saturation solubility of the salt AB be "s" mol/L. Since the salt is a 1:1 salt, at equilibrium, [A\(^{+}\)] = [B\(^{-}\)] = s. Now we can set up the equation using the K\(_{sp}\) value: K\(_{sp}\) = (s)(s) = s\(^{2}\) To calculate the saturation solubility, solve the equation for s: s = \(\sqrt{K_{sp}}\) By plugging in the K\(_{sp}\) value and taking the square root, we can quickly determine the saturation solubility of the sparingly soluble 1:1 salt at 25°C.

Key concepts

Solubility Product Constant (Ksp)

When we try to dissolve a sparingly soluble salt in water, a dynamic equilibrium between the solid phase and its ions in solution is established. The equilibrium constant for this reaction is known as the solubility product constant, designated as K_sp. This numerical value is crucial for predicting solubility and helps chemists determine the concentration of ions that will exist in a saturated solution.

For a simple salt AB that dissociates into one cation A⁺ and one anion B^-, the K_sp expression is K_sp = [A⁺][B^-]. It's important to note that K_sp is specific to each salt and depends on temperature. Typically, its value is found through experimentation and tabulated for reference. Understanding K_sp is fundamental because it allows us to calculate the maximum amount of a salt that can dissolve in water before it becomes a saturated solution.

Dissolution Equilibrium

Dissolution equilibrium occurs when a reversible reaction establishes between the undissolved solute and the ions in a solution. Speaking about sparingly soluble salts, they have a characteristic of reaching this state quickly due to their low solubility. At equilibrium, the rate at which the solid solute dissolves to form its respective ions is equal to the rate at which these ions combine to form the solid solute.

Representing this concept with the generic salt AB, we use the chemical equation AB(s) <==> A⁺(aq) + B^-(aq). Here, '(s)' denotes the solid phase, and '(aq)' denotes the aqueous or dissolved ions. In a saturated solution, once equilibrium is reached, no more solute will dissolve, and this is indicated by constant concentrations of the dissolved ions.

Equilibrium Concentration

The equilibrium concentration refers to the amount of solute that is in solution at equilibrium. For a 1:1 sparingly soluble salt, such as the generic salt AB, the equilibrium concentrations of the cation A⁺ and the anion B^- are equal, due to their stoichiometric relationship in the dissociation equation. If the saturation solubility of the salt is represented as 's' molarity (mol/L), it means that both [A⁺] and [B^-] are also 's'.

To find the individual ion concentrations in a saturated solution from K_sp, one can assume that the concentration of each ion is the square root of the K_sp value for salts with a 1:1 ratio between cation and anion, which simplifies calculations. Comprehending the concept of equilibrium concentration is imperative for predicting the outcome of reactions and for understanding how concentrations can be manipulated in various chemical processes.

Sparingly Soluble Salts

Sparingly soluble salts are those that have very low solubility in water – so low that the undissolved substance is noticeable and the solution is almost at the saturation point. These types of salts are distinct because they often reach a state of dynamic equilibrium where the rate of dissolution and the rate of precipitation are equal, leading to a constant concentration of ions in solution.

Understanding the nature of sparingly soluble salts is important when discussing the concept of saturation solubility. In general, compounds with K_sp values less than 1 x 10^-6 are considered sparingly soluble. These compounds pose unique challenges and opportunities in chemical synthesis, water purification, and pharmaceutical formulation. Recognizing the solubility characteristics of such salts is key in aligning with appropriate methods for their analysis and usage.