Problem 50 - Chemistry The Central Science In Si Units

Question

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, \(4 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\). (a) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2},\) what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+} ?\)

Step-by-Step Solution

Answer

(a) MA has the larger Ksp value. (b) Both have equal \(\mathrm{M}^{2+}\) concentration. (c) The equilibrium \(\mathrm{M}^{2+}\) concentration is \(4 \times 10^{-4} \text{ mol/L}\).

Step 1: Understand What Solubility Means

The solubility given is the maximum amount of solute (salt in this case) that can dissolve in a given volume of solvent to form a saturated solution. Here, both salts, MA and MZ2, have a solubility of \(4 \times 10^{-4} \text{ mol/L}\).

Step 2: Write Solubility Equations

For MA dissociation, the equation is \(\text{MA} \rightleftharpoons \text{M}^{2+} + \text{A}^{2-}\). For MZ2 dissociation, it is \(\text{MZ}_2 \rightleftharpoons \text{M}^{2+} + 2\text{Z}^-\).

Step 3: Solubility Product Constants Calculation

The solubility product (Ksp) is the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation. For MA, \(\text{K}_{\text{sp}} = [\text{M}^{2+}] [\text{A}^{2-}] = (4 \times 10^{-4})(4 \times 10^{-4}) = 1.6 \times 10^{-7}\). For MZ2, \(\text{K}_{\text{sp}} = [\text{M}^{2+}] [\text{Z}^-]^2 = (4 \times 10^{-4})(8 \times 10^{-4})^2 = 2.56 \times 10^{-10}\).

Step 4: Compare Solubility Products

Comparing the calculated Ksp values, \(\text{Ksp}_{\text{MA}} = 1.6 \times 10^{-7}\) and \(\text{Ksp}_{\text{MZ}_2} = 2.56 \times 10^{-10}\). The solubility product of MA is larger.

Step 5: Calculate M²⁺ Concentration in Saturated Solutions

For MA, the concentration of \(\text{M}^{2+}\) is \(4 \times 10^{-4} \text{ mol/L}\) and for MZ2 it is also \(4 \times 10^{-4} \text{ mol/L} \). Hence, they have equal concentrations of \(\text{M}^{2+}\).

Step 6: Determine Equilibrium Concentration After Mixing

Mixing equal volumes will dilute the concentration of \(\text{M}^{2+}\) from each source. After mixing, the dilution factor is 2, so \([\text{M}^{2+}] = \frac{4 \times 10^{-4}}{2} + \frac{4 \times 10^{-4}}{2} = 4 \times 10^{-4} \text{ mol/L}\).

Key Concepts

Solubility Equations

In chemistry, understanding solubility equations is essential for predicting how substances will dissolve and interact in solutions. When a salt dissolves in water, it dissociates into its constituent ions. This process can be represented using chemical equations, known as solubility equations.

Let's consider two salts from the exercise: MA and MZ\(_2\). Both salts dissociate in water to form ions. For MA, the equation is:

• \( \text{MA} \rightleftharpoons \text{M}^{2+} + \text{A}^{2-} \)

For MZ\(_2\), the dissociation equation is slightly different:

• \( \text{MZ}_{2} \rightleftharpoons \text{M}^{2+} + 2\text{Z}^{-} \)

This indicates that for every mole of MZ\(_2\) that dissolves, it produces one mole of \(\text{M}^{2+}\) ions and two moles of \(\text{Z}^-\) ions. These equations help us understand how the ratio of ions is determined by the chemical formula of the salt. Knowing these equations allows you to calculate concentrations and other important properties of the solution.

Equilibrium Concentration

When a salt dissolves in water, it reaches a state known as dynamic equilibrium. At this point, the rate at which the solid salt dissolves equals the rate at which the ions precipitate out of solution.

For the given salts MA and MZ\(_2\), equilibrium concentrations are determined by their solubility and the dissociation equations:

• For MA: The equilibrium concentration for \(|\text{M}^{2+}|\) ions is the same as the solubility value, \(4 \times 10^{-4} \text{ mol/L}\).
• For MZ\(_2\): Although the initial molar solubility is the same, the presence of additional ions affects the equilibrium. It reaches the same maximum \(|\text{M}^{2+}|\) concentration, due to equal solubility rates.

The calculations show that both salts achieve the same equilibrium concentration for the \(|\text{M}^{2+}|\) ions under these conditions, signifying that equilibrium reflects a balance that not only depends on how much of a substance dissolves, but also on the stoichiometry of its dissociation.

Molar Concentrations

Molar concentrations play a crucial role in understanding solubility and chemical equilibriums in solutions. It measures the amount of a solute, in moles, per liter of solution.

To determine molar concentrations from solubility, you need to consider the dissociation of each compound. For instance, when calculating the solubility product constant (Ksp):

• For MA, both \(|\text{M}^{2+}|\) and \(|\text{A}^{2-}|\) have the same concentration in solution, \(4 \times 10^{-4} \text{ mol/L}\).
• For MZ\(_2\), the concentration of \(|\text{Z}^{-}|\) is twice that of \(|\text{M}^{2+}|\), reflecting the formula \(\left[\text{Z}^{-}\right] = 2 \times 4 \times 10^{-4} \text{ mol/L}\).

These values directly affect the calculation of Ksp. For MA, \(\text{K}_{\text{sp}}\) is determined by multiplying the concentrations of \(|\text{M}^{2+}|\) and \(|\text{A}^{2-}|\), while for MZ\(_2\), \(\text{K}_{\text{sp}}\) combines \(|\text{M}^{2+}|\) and \([\text{Z}^-]^2\). Understanding these relationships can help predict how different solutions will behave and allow you to solve similar problems with confidence.