Problem 42
Question
Calculate the solubility of \(\mathrm{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a \(\mathrm{pH}\) of 11.00.
Step-by-Step Solution
Verified Answer
The solubility of Co(OH)₂ in a buffered solution with a pH of 11.00 is calculated using the given Ksp value and the concentration of OH⁻ ions. Using the dissolution of Co(OH)₂ in water, the balanced chemical equation, and the Ksp expression, we find the concentration of Co²⁺ ions. After converting the given pH to the concentration of OH⁻ ions and substituting the given Ksp and [OH⁻] into the Ksp expression, we solve for the concentration of Co²⁺ ions. The solubility of Co(OH)₂ in the buffered solution is found to be 2.5 × 10⁻¹⁰ M.
1Step 1: Write the balanced chemical equation
The dissolution of Co(OH)2 in water can be represented by the following balanced chemical equation:
Co(OH)2 (s) ↔ Co2+ (aq) + 2 OH- (aq)
2Step 2: Write the expression for Ksp
The solubility product constant (Ksp) for the dissolution of Co(OH)2 is given by:
Ksp = [Co²⁺] × [OH⁻]²
where [Co²⁺] and [OH⁻] are the equilibrium concentrations of Co2+ and OH- ions, respectively. We are given that Ksp = 2.5 × 10⁻¹⁶.
3Step 3: Convert the given pH to [OH⁻] concentration
We are given the pH of the buffered solution as 11.00. To find the concentration of OH- ions, we will first find the pOH:
pOH = 14 - pH = 14 - 11 = 3
Now, we can calculate the concentration of OH- ions using the pOH:
[OH⁻] = 10^(-pOH) = 10⁻³ M
4Step 4: Substitute the given Ksp and [OH⁻] into the Ksp expression
Now that we have the concentration of OH- ions, we can substitute it into the Ksp expression:
2.5 × 10⁻¹⁶ = [Co²⁺] × (10⁻³)²
2.5 × 10⁻¹⁶ = [Co²⁺] × 10⁻⁶
5Step 5: Solve for the concentration of Co²⁺ ions
Now we will solve for the equilibrium concentration of Co²⁺ ions:
[Co²⁺] = (2.5 × 10⁻¹⁶) / 10⁻⁶
[Co²⁺] = 2.5 × 10⁻¹⁰ M
6Step 6: Calculate the solubility of Co(OH)2
The solubility of Co(OH)2 in the given buffered solution is equal to the equilibrium concentration of Co²⁺ ions:
Solubility of Co(OH)2 = 2.5 × 10⁻¹⁰ M
Key Concepts
KspChemical EquilibriumpH and pOHBuffer Solutions
Ksp
When discussing solubility, the solubility product constant, or Ksp, is a key concept. It's a measure of how much of a compound can dissolve in a solution before it begins to precipitate. Imagine Ksp as a balance between the solid and its ions in solution. For cobalt(II) hydroxide, the equation is expressed as:\[ \text{Co(OH)}_2 (s) \rightleftharpoons \text{Co}^{2+} (aq) + 2\text{OH}^{-} (aq) \]The Ksp equation becomes:\[ K_{sp} = [\text{Co}^{2+}] \times [\text{OH}^-]^2 \]This helps us find the equilibrium concentrations of ions, acting as a snapshot of how much dissolves before reaching saturation. It's fascinating how small changes in the solution can drastically shift the equilibrium.
Chemical Equilibrium
Chemical equilibrium is a dynamic state where the concentration of reactants and products remains constant over time. For solubility, equilibrium means the rate at which Co(OH)₂ dissolves is equal to the rate at which it crystallizes.
Here's how it works:
- The system balances between the dissolved ions and the undissolved solid.
- Even at equilibrium, reactions do not stop; they just proceed at equal rates.
- The concentrations of ions are governed by the Ksp value, setting the maximum solubility under given conditions.
pH and pOH
Understanding pH and pOH is essential when dealing with buffered solutions, as these influence solubility. pH represents the acidity, and pOH the alkalinity, both summing to 14 in a solution:\[ \text{pH} + \text{pOH} = 14 \]For the solution with a pH of 11, we find the pOH is 3. This reveals important details:
- A high pH (or low pOH) indicates a basic environment.
- Such conditions shift equilibrium, affecting how many ions the Co(OH)₂ can release.
- In this instance, more OH⁻ ions are already present, impacting how much cobalt can dissolve.
Buffer Solutions
Buffer solutions maintain a stable pH, crucial for precise solubility calculations. They are mixtures of weak acids and their conjugate bases, or vice versa. Here's what they achieve:
- Resist changes in pH, even when strong acids or bases are added.
- Provide a controlled environment to study reactions like Co(OH)₂ solubility.
- Allow us to assume certain concentrations, like the OH⁻ found previously, as constant.
Other exercises in this chapter
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