Question

Solubility of a \(\mathrm{M}_{2} \mathrm{~S}\) salt is \(3.5 \times 10^{-6}\) then find out solubility product: (a) \(1.7 \times 10^{-6}\) (b) \(1.7 \times 10^{-16}\) (c) \(1.7 \times 10^{-18}\) (d) \(1.7 \times 10^{-12}\)

Short answer

The solubility product \( K_{sp} \) is \( 1.7 \times 10^{-16} \).

Step-by-step solution

Understand the Solubility Model

The given salt is \( \mathrm{M}_{2} \mathrm{~S} \), meaning it will dissociate in water as follows: \[ \mathrm{M}_{2} \mathrm{~S} \rightarrow 2\mathrm{M}^{+} + \mathrm{S}^{2-} \] Thus, when the salt dissolves, it produces 2 moles of \( \mathrm{M}^{+} \) ions and 1 mole of \( \mathrm{S}^{2-} \) ions.

Define and Calculate Ion Concentrations

If the solubility of \( \mathrm{M}_{2} \mathrm{~S} \) is \( 3.5 \times 10^{-6} \) M, the concentration of \( \mathrm{S}^{2-} \) ions is \( 3.5 \times 10^{-6} \) M, and the concentration of \( \mathrm{M}^{+} \) ions is \( 2 \times (3.5 \times 10^{-6}) = 7.0 \times 10^{-6} \) M.

Write the Expression for Solubility Product (Ksp)

The solubility product \( K_{sp} \) is expressed as: \[ K_{sp} = [\mathrm{M}^{+}]^2 [\mathrm{S}^{2-}] \] Plugging in the values: \[ K_{sp} = (7.0 \times 10^{-6})^2 \times (3.5 \times 10^{-6}) \]

Perform the Calculation

Calculate \( K_{sp} \) using the values from Step 3: \[ (7.0 \times 10^{-6})^2 = 49 \times 10^{-12} = 4.9 \times 10^{-11} \] \[ K_{sp} = 4.9 \times 10^{-11} \times 3.5 \times 10^{-6} = 1.715 \times 10^{-16} \]

Key concepts

Solubility Calculation

Solubility calculation is key when you want to know the extent to which a compound dissolves in a solvent. Think of it as measuring how much of a salt, like \( \mathrm{M}_{2} \mathrm{~S} \), can be dissolved in water to form its ions. In our exercise, the solubility given is \(3.5 \times 10^{-6}\, \text{M}\). This value tells us the maximum concentration of the dissolved salt in water at equilibrium.

To perform a solubility calculation, you often begin with this solubility value to determine the concentrations of the ions formed by the compound in solution. For \( \mathrm{M}_{2} \mathrm{~S} \), each mole dissolves to give \(2\) \( \mathrm{M}^{+} \) ions and \(1\) \( \mathrm{S}^{2-} \) ion. These concentrations are crucial for later steps, such as calculating the solubility product, \( K_{sp} \).

Thus, solubility calculation acts as a starting point for understanding the complete dissolution process for various salts, providing the necessary information to calculate the solubility product.

Ionic Dissociation

Ionic dissociation is the process by which an ionic compound like \( \mathrm{M}_{2} \mathrm{~S} \) separates into its individual ions when dissolved in water. This process is crucial for solubility calculations because it defines how the compound breaks apart.

In the case of \( \mathrm{M}_{2} \mathrm{~S} \), when it dissolves, it dissociates as:

• One mole of \( \mathrm{M}_{2} \mathrm{~S} \) results in two moles of \( \mathrm{M}^{+} \) ions, and
• One mole of \( \mathrm{S}^{2-} \) ions.

This means the dissociation creates a specific ratio of ions in the solution. For every mole of \( \mathrm{M}_{2} \mathrm{~S} \) that dissolves, two moles of \( \mathrm{M}^{+} \) are produced, which doubles the concentration needed to be considered in subsequent calculations.

Understanding ionic dissociation helps clarify how ions will interact in solution, influencing properties like conductivity and the ability to undergo chemical reactions. It is a foundational concept in chemistry problem-solving and integral to predicting the behavior of salts when they dissolve.

Chemistry Problem-Solving

Chemistry problem-solving often requires a multi-step approach. This methodical progression is key to solving problems like the one involving the solubility product. To successfully approach chemistry problems:

• Break down the problem into smaller, manageable steps.
• Clearly understand the concepts involved at each stage, such as ionic dissociation and solubility calculations.

This step-by-step analysis helps in managing complexity by addressing each part individually.

In our exercise with \( \mathrm{M}_{2} \mathrm{~S} \), the steps include understanding the compound's dissociation, identifying ion concentrations, and using these to calculate the \( K_{sp} \). Each of these steps builds upon the previous one, creating a clear path from the problem's initial conditions to its final solution.

Using a structured approach when tackling chemistry problems not only provides clarity and direction but also significantly reduces the chances of errors, leading to successful and accurate results. This systematic strategy is what makes complex chemistry concepts accessible and solvable.

Ksp Expression

The \( K_{sp} \) expression, or solubility product constant, is a crucial concept in understanding the solubility in chemical solutions. It describes the product of the concentrations of the ions of a dissolved compound, each raised to the power of its coefficient in the balanced chemical equation.

For the dissociation of \( \mathrm{M}_{2} \mathrm{~S} \), the \( K_{sp} \) expression is given by:\[ K_{sp} = [\mathrm{M}^{+}]^2 [\mathrm{S}^{2-}] \]This expression reflects the equilibrium conditions of the dissolved ions in solution. To find \( K_{sp} \), we use the concentrations calculated from solubility, where \( [\mathrm{M}^{+}] = 7.0 \times 10^{-6} \text{ M} \) and \( [\mathrm{S}^{2-}] = 3.5 \times 10^{-6} \text{ M} \).

The solubility product tells us how likely a salt is to dissolve in water, providing a quantitative measure of solubility. Understanding the \( K_{sp} \) helps predict whether precipitation will occur in a solution when different ions are present. It is a key concept in analyzing and predicting chemical equilibria in various scenarios.