Problem 50 - Chemistry: The Central Science

Question

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\) are 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 \(\mathrm{M}^{2+}\) ? (c) If you added an equal volume of a solution saturated in \(\mathrm{MA}\) 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 a larger numerical value for the solubility product constant. (b) In a saturated solution of each salt in water, both salts have the same concentration of M²⁺. (c) The equilibrium concentration of the cation, M²⁺, will be \(8 \times 10^{-4}\) mol/L after mixing equal volumes of the saturated solutions.

Step 1: 1. Write the Dissolution Equations

Since we are given two salts, MA and MZ₂, let's write the dissolution equations for them: For MA: \( \mathrm{MA \rightleftharpoons M^{2+} + A^{-}} \) For MZ₂: \( \mathrm{MZ_{2} \rightleftharpoons M^{2+} + 2Z^{-}} \)

Step 2: 2. Write the Solubility Product Expressions

Now, we will write the solubility product expressions for the salts: For MA: \( K_{\mathrm{sp,MA}} = [\mathrm{M^{2+}}][\mathrm{A^{-}}] \) For MZ₂: \( K_{\mathrm{sp,MZ_{2}}} = [\mathrm{M^{2+}}][\mathrm{Z^{-}}]^2 \)

Step 3: 3. Substitute the Solubilities

Given the solubilities of the salts are the same (\(4 \times 10^{-4}\)), we will substitute them into the solubility product expressions: For MA: \( K_{\mathrm{sp,MA}} = (4 \times 10^{-4})(4 \times 10^{-4}) \) For MZ₂: \( K_{\mathrm{sp,MZ_{2}}} = (4 \times 10^{-4})(2(4 \times 10^{-4}))^2 \)

Step 4: 4. Calculate the Solubility Product Constants

Now, let's calculate the solubility product constants: For MA: \(K_{\mathrm{sp,MA}} = (4 \times 10^{-4})^2 = 16 \times 10^{-8}\) For MZ₂: \(K_{\mathrm{sp,MZ_{2}}} = (4 \times 10^{-4})(2 \times 4 \times 10^{-4})^2 = (4 \times 10^{-4})( 8 \times 10^{-4})^2 = 64 \times 10^{-12}\)

Step 5: 5. Compare the Solubility Product Constants

Comparing the two solubility product constants: \(K_{\mathrm{sp,MA}} = 16 \times 10^{-8}\) \(K_{\mathrm{sp,MZ_{2}}} = 64 \times 10^{-12}\) (a) MA has a larger numerical value for the solubility product constant.

Step 6: 6. Determine the Higher Concentration of M²⁺

Given that both salts have the same solubility of \(4 \times 10^{-4}\) mol/L: (b) In a saturated solution of each salt in water, both salts have the same concentration of M²⁺.

Step 7: 7. Find the Equilibrium Concentration M²⁺ After Mixing

(c) When equal volumes of saturated solutions of both salts are mixed, the concentration of M²⁺ will double because both salts contribute the same concentration. Thus, the new concentration of M²⁺ will be: \(2 \times (4 \times 10^{-4}) = 8 \times 10^{-4}\) The equilibrium concentration of the cation, M²⁺, will be \(8 \times 10^{-4}\) mol/L after mixing equal volumes of the saturated solutions.

Key Concepts

Understanding Solubility

Solubility refers to the ability of a substance to dissolve in a solvent, such as water. In chemistry, it is defined as the maximum amount of a solute that can be dissolved in a solvent at a specific temperature to form a saturated solution. To visualize this, imagine sugar dissolving in water; there is a limit to how much can dissolve before additional sugar just settles at the bottom.

Solubility is influenced by various factors like temperature, pressure, and the nature of the solute and solvent. For instance, most solid solutes increase in solubility with temperature, while gases typically decrease. Water's polarity allows it to dissolve many ionic compounds by separating the ions and surrounding them, this interaction is key for the dissolution process. Still, not all ionic compounds dissolve well; such compounds are called 'slightly soluble' or 'sparingly soluble', leading us to the concept of the solubility product constant.

Navigating Ksp Calculations

The solubility product constant, or Ksp, is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. It represents how much a compound dissociates to form ions in solution at equilibrium. The smaller the Ksp, the less soluble the compound is.

Let's consider the exercise above, where we determine Ksp for the salts MA and MZ₂. Ksp calculations involve setting up an equation that is equal to the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced dissolution equation. For instance, the Ksp for a generic salt AB₂ would be written as \[K_{\text{sp}} = [A^+][B^-]^2\]. In the given exercise, the correct substitution of solubility values into the Ksp expressions and squaring or cubing them according to stoichiometry is the crux of determining the numerical value of each Ksp.

Determining Equilibrium Concentration

Equilibrium concentration is the concentration of each ion or molecule in a system at equilibrium. At this point, the rate of the forward reaction (dissolution) equals the rate of the backward reaction (precipitation), so the net concentration of reactants and products remains constant over time.

In the case of mixing two solutions of sparingly soluble salts with a common ion, like M²⁺, the initial concentrations before mixing are simply the solubility of each salt. After mixing, if volumes are equal, the concentration of common ions simply adds up, as shown in step 7 of the solution. It's important not just to understand how to calculate this, but why it happens: the process maintains the principles of conservation of mass—ions are neither created nor destroyed when solutions are mixed.

Mastering Dissolution Equations

Dissolution equations are written representations of the process by which a solute dissolves in a solvent to form a solution. In the context of sparingly soluble salts, it is essential to know the stoichiometry of the reaction: how many ions of each type are produced when the solid dissolves.

In the exercise, we see two dissolution equations, one for salt MA and another for MZ₂. Writing these equations is the first step in understanding the dissolution process, as they allow us to set up the solubility product expression. Correctly balancing these equations is crucial; incorrectly balanced equations can lead to errors in Ksp calculations and subsequently, in determining the equilibrium concentrations. After establishing the balanced equation, the stoichiometry is then used to determine the Ksp expression and eventually the solubility product constant.