Problem 116 - Chemistry

Question

The salt MX has a solubility of \(3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\) in a solution with \(\mathrm{pH}=0.000 .\) If \(K_{\mathrm{a}}\) for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

Step-by-Step Solution

Answer

The solubility product constant (Ksp) for the salt MX can be calculated using the given solubility, pH, and Ka values. After determining the concentrations of ions in the solution, the Ksp expression for MX is [M+][X-], which yields Ksp = \((3.17 \times 10^{-8})(3.17 \times 10^{-8}) = 1.01 \times 10^{-15}\). Therefore, the Ksp value for MX is \(1.01 \times 10^{-15}\).

Step 1: Write the Dissolution and Ionization Equations

First, write the equation for the dissolution of MX, and then the equation for the ionization of the weak acid HX: Dissolution equation: \(MX → M^{+} + X^{-}\) Ionization equation: \(HX \rightleftharpoons H^{+} + X^{-}\) Since we know the solubility of MX and the pH of the solution, we can use this information to find the concentrations of the ions.

Step 2: Determine the Concentrations of Ions in Solution

To find the concentrations of the ions, let x be the equilibrium concentration of H+. Then, [x] = \(10^{-pH} = 10^{-0} = 1\) The concentration of X- ions will be equal to the solubility of MX: [X-] = \(3.17 \times 10^{-8}\) mol/L Since the system is at equilibrium, the concentration of HX will also be \(3.17 \times 10^{-8}\) mol/L.

Step 3: Find the Concentration of M+ Ions

At equilibrium, the concentration of M+ ions will be the same as the concentration of X- ions: [M+] = [X-] = \(3.17 \times 10^{-8}\) mol/L

Step 4: Calculate the Concentration of HX Using Ka

With the given Ka value, we can now create an equation to find the concentration of HX in equilibrium: Ka = \(\frac{[H^{+}][X^{-}]}{[HX]}\) Plug in the known values to determine the concentration of [HX]: \(1.00 \times 10^{-15} = \frac{(1)(3.17 \times 10^{-8})}{[HX]}\) Solve for [HX]: [HX] = \(\frac{3.17 \times 10^{-8}}{1.00 \times 10^{-15}}\) = \(3.17 \times 10^{7}\) mol/L

Step 5: Calculate the Ksp of MX

Now we have all the necessary information to find the Ksp. Write the Ksp expression for MX: Ksp = [M+][X-] Plug in the known concentrations: Ksp = \((3.17 \times 10^{-8})(3.17 \times 10^{-8})\) Calculate the Ksp value for MX: Ksp = \(1.01 \times 10^{-15}\) Thus, the solubility product constant (Ksp) for MX is \(1.01 \times 10^{-15}\).

Key Concepts

Ksp calculation

Calculating the solubility product constant, or Ksp, is a key step when working with dissolution equilibria for slightly soluble salts. In this scenario, the salt MX dissolves in water to form ions. The equation for this process is: \( MX \rightarrow M^{+} + X^{-} \). From the given solubility of MX as \(3.17 \times 10^{-8} \; \text{mol/L}\), we can determine the concentration of each ion in the solution because it dissociates equally into \( M^{+} \) and \( X^{-} \) ions. This means that both \([M^+]\) and \([X^-]\) are equal to \(3.17 \times 10^{-8} \; \text{mol/L}\).

To find the Ksp of MX, we use the expression:

• \( K_{sp} = [M^+][X^-] \)

Substituting the values into the expression gives: \( K_{sp} = (3.17 \times 10^{-8})(3.17 \times 10^{-8}) \). This calculation results in a Ksp value of \(1.01 \times 10^{-15}\). This value reflects the very low solubility of MX in the solution.

dissolution equilibrium

Dissolution equilibrium refers to the balance reached when a salt dissolves in water to form ions. In this case, MX is a salt that dissociates into its constituent ions, \( M^{+} \) and \( X^{-} \), in solution. At equilibrium, the process of dissolution (MX dissolving) and precipitation (ions recombining to form MX) occurs at equal rates.

In our given exercise, the ion concentrations reach a stable state where they do not change further, unless the system's conditions (such as temperature or pressure) are altered.

• The dissolution equation: \( MX \rightarrow M^{+} + X^{-} \) helps in understanding how equilibrium is maintained.
• The concentration of ions present at equilibrium is determined by the salt's initial solubility.

As the dissolution occurs in a solution with \( \text{pH} = 0\), indicating a highly acidic environment, the significant concentration of \([H^+]\) also plays a role in determining the concentration of ions derived from weak acid ionization.

weak acid ionization

Weak acid ionization is important in solutions where the weak acid interacts with the dissolved ions of a salt. Here, HX is the weak acid undergoing ionization, represented as \( HX \rightleftharpoons H^{+} + X^{-} \). At equilibrium, the concentrations of the weak acid, its ions, and other contributing ions from the solute govern the overall balance.

Recall that:

• The \(K_{a}\) value provides insight into how an acid dissociates. For HX, a given \(K_{a} = 1.00 \times 10^{-15}\) shows very limited ionization.
• The concentration of \([H^+]\) derived from the system's pH directly affects ionization equilibrium.

This equation, \( K_a = \frac{[H^+][X^-]}{[HX]} \), allows for calculation of ion concentrations. Solving for \([HX]\) provides an understanding of how much HX remains un-ionized versus how much has dissociated into ions. Given the low \(Ka\) value and the calculated concentrations, it indicates HX mostly remains in its undissociated form within this acidic solution. This reflects the interplay between weak acid ionization and the solubility of salts in the given environment.