Question

The molar solubility (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) of a sparingly soluble salt \(\mathrm{MX}_{4}\) is 's'. The corresponding solubility product is Ksp.s is given in terms of Ksp by the relation: (a) \(\mathrm{s}=(\mathrm{Ksp} / 256)^{1 / 5}\) (b) \(\mathrm{s}=(128 \mathrm{Ksp})^{1 / 4}\) (c) \(\mathrm{s}=(\mathrm{Ksp} / 128)^{1 / 4}\) (c) \(\mathrm{s}=(256 \mathrm{Ksp})^{1 / 5}\)

Short answer

The correct relation is \( s = \left(\frac{K_{sp}}{256}\right)^{1/5} \).

Step-by-step solution

Write the dissociation equation

The salt \( \text{MX}_4 \) dissociates in water as follows: \[ \text{MX}_4 \leftrightarrow \text{M}^{4+} + 4\text{X}^- \]

Define the expression for Ksp

The solubility product (\( K_\text{sp} \)) is expressed in terms of the concentrations of ions at equilibrium. For \( \text{MX}_4 \) dissociating to \( \text{M}^{4+} \) and \( 4\text{X}^- \), \[ K_\text{sp} = [\text{M}^{4+}][\text{X}^-]^4 \]

Express ion concentrations in terms of 's'

Let the molar solubility of \( \text{MX}_4 \) be \( s \). At equilibrium, \[ [\text{M}^{4+}] = s \quad \text{and} \quad [\text{X}^-] = 4s \] These are based on the stoichiometry of the dissociation reaction.

Substitute ion concentrations in Ksp expression

Substitute \( [\text{M}^{4+}] = s \) and \( [\text{X}^-] = 4s \) into the \( K_\text{sp} \) expression: \[ K_\text{sp} = s (4s)^4 = s \times 256s^4 = 256s^5 \]

Solve for 's' in terms of Ksp

Rearrange the equation \( K_\text{sp} = 256s^5 \) to find \( s \). Divide both sides by 256 and take the fifth root: \[ s = \left(\frac{K_\text{sp}}{256}\right)^{1/5} \] This simplifies to the expression for molar solubility.

Key concepts

Molar Solubility

Molar solubility is a measure of the number of moles of a solute, such as a sparingly soluble salt, that can dissolve in a liter of solution to reach saturation. It helps us understand how much solute will dissolve in solution before reaching equilibrium. The value of molar solubility is particularly important in predicting how a salt behaves in a solution, and it is denoted usually by 's'.
One of the key applications of molar solubility is in calculating the solubility product ( K_ ext{sp} ), which gives us insight into the solubility and the dissolution process of salts.
By knowing K_ ext{sp} and the type of salt, we can determine the molar solubility and understand the extent to which a salt can dissolve in water. This value is crucial for many scientific investigations, ranging from environmental science to pharmaceuticals.

Sparingly Soluble Salts

Sparingly soluble salts, as the name suggests, are salts that do not dissolve well in water. They have very low solubility, meaning only a small amount of the salt can dissolve before the solution becomes saturated. Examples include salts like MX_4 , which are often studied using their solubility products.
Understanding the behavior of these salts is important in several fields, particularly in predicting mineral formation in geology and controlling precipitation reactions in chemistry.
The solubility product ( K_ ext{sp} ) is a key indicator when studying sparingly soluble salts, as it allows chemists to predict whether a precipitate will form when solutions are mixed.

Dissociation Equation

The dissociation equation is a representation of how a solid salt breaks down into its constituent ions when it dissolves in water. For the salt MX_4, the dissociation can be expressed as:
\[ \text{MX}_4 \leftrightarrow \text{M}^{4+} + 4\text{X}^- \]
This equation shows how one formula unit of the salt produces one M^{4+} ion and four X^- ions. It is important for understanding the stoichiometry of the reaction and is crucial when calculating K_ ext{sp}.
When we know how a salt dissociates, we can better predict the concentration of ions in solution and subsequently the behavior of the salt in various chemical environments.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of ions in a solution when a dissolved salt has reached a point where no more salt can dissolve, i.e., saturation. In the context of the dissociation of MX_4, the equilibrium concentrations are noted as [s] for M^{4+} and [4s] for X^-.
This concept is essential for calculating the solubility product (K_ ext{sp}) because K_ ext{sp} is inherently dependent on the concentrations of each ion at equilibrium:
\[ K_\text{sp} = [\text{M}^{4+}][\text{X}^-]^4 \]
By plugging the equilibrium concentrations into the K_ ext{sp} equation, we determine the solubility of the salt in a given solution, which is critical for solving various real-world problems involving solubility and precipitation.