Question

The \(\mathrm{pH}\) of a saturated solution of a metal hydroxide MOH is 9.68 . Calculate the \(K_{s p}\) for this compound.

Short answer

\(K_{sp} = 5.25 \times 10^{-9}\).

Step-by-step solution

Understanding the Problem

We are given that the pH of a saturated solution of a metal hydroxide, MOH, is 9.68. We need to find the solubility product constant, \(K_{sp}\), of the hydroxide.

Calculate pOH

Since \(\text{pH} + \text{pOH} = 14\), we can find \(\text{pOH}\) by subtracting the given pH from 14. Thus, \(\text{pOH} = 14 - 9.68 = 4.32\).

Determine [OH⁻] Concentration

The hydroxide ion concentration, \([\text{OH}^-]\), can be calculated using the formula \([\text{OH}^-] = 10^{-\text{pOH}}\). Therefore, \([\text{OH}^-] = 10^{-4.32}\).

Write the Dissociation Reaction

The dissociation of MOH in water is \(\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-\). For each mole of MOH that dissociates, one mole of \([\text{OH}^-]\) and \([\text{M}^+]\) is formed. Thus, \([\text{M}^+] = [\text{OH}^-]\).

Express Ksp in terms of [OH⁻]

The solubility product constant, \(K_{sp}\), is given by \(K_{sp} = [\text{M}^+][\text{OH}^-]\). Since \([\text{M}^+] = [\text{OH}^-]\), \(K_{sp} = [\text{OH}^-]^2\).

Calculate Ksp

Substitute the hydroxide ion concentration into the expression for \(K_{sp}\): \(K_{sp} = (10^{-4.32})^2\). Solve this to find \(K_{sp}\).

Key concepts

pH Calculation

The concept of pH is essential in chemistry, as it helps us understand the acidity or basicity of a solution. pH is a measure of the hydrogen ion concentration. It's calculated using the formula:

• \[\text{pH} = -\log[H^+]\]

This exercise provides the pH of a solution, which is 9.68. Knowing the pH allows us to determine the pOH because the sum of pH and pOH in an aqueous solution at 25°C is always 14. This relationship helps us move between pH and pOH easily, aiding in further calculations.
To find the pOH from the pH, use the formula:

• \[\text{pOH} = 14 - \text{pH}\]

In this case, the pOH is 4.32.

Dissociation Reaction

A dissociation reaction is a process where a compound breaks down into its ions. In our exercise, the compound MOH dissociates in water. The dissociation equation is:

• \[\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-\]

This equation shows that when one molecule of MOH dissociates, one metal ion \(\text{M}^+\) and one hydroxide ion \(\text{OH}^-\) are produced.
Dissociation is key to understanding reactions in solutions, where chemical equilibrium establishes how much of the compound breaks into ions. Knowing the balance of ions guides us in calculating the solubility product constant \(K_{sp}\). It ties directly into understanding changes in concentrations.

Hydroxide Ion Concentration

Hydroxide ion concentration \([OH^-]\) impacts pH and the solubility product constant. To find \([OH^-]\), use the pOH determined from pH calculations. The formula is:

• \[[\text{OH}^-] = 10^{-\text{pOH}}\]

With a pOH of 4.32, the hydroxide ion concentration in this exercise is given by \(10^{-4.32}\).
This concentration tells us the amount of hydroxide ions present in the solution, which is crucial for assessing the solution's basicity. It also helps calculate \(K_{sp}\), allowing us to see how soluble the compound really is.

Chemical Equilibrium

In a chemical reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. For our dissociation example, chemical equilibrium describes how the concentrations of MOH and its ions stabilize over time.

• At equilibrium in a saturated solution of MOH:
  • Concentration of \(\text{M}^+\) equals concentration of \(\text{OH}^-\).
  • \(K_{sp} = [\text{M}^+][\text{OH}^-] = [\text{OH}^-]^2\)

The calculation of \(K_{sp}\) helps predict solubility, which indicates the level to which a solute will dissolve. It's essential in understanding how much of the solid will remain in solution at equilibrium.
By calculating \(K_{sp}\), students gain insight into the balance between dissolved ions and undissolved solid, a core principle of chemical equilibrium and solubility.