Problem 94 - Chemistry An Atoms First Approach

Question

Calculate the concentration of an aqueous \(\operatorname{Sr}(\mathrm{OH})_{2}\) that has \(\mathrm{pH}=10.50\)

Step-by-Step Solution

Answer

The concentration of the aqueous Sr(OH)₂ solution with a pH of 10.50 is approximately \(1.59 \times 10^{-4}\) mol/L.

Step 1: Find the hydrogen ion concentration

We are given the pH value of the solution, which is 10.50. To find the H⁺ concentration, we can use the equation: pH = -log[H⁺] where pH is the pH value and [H⁺] is the H⁺ concentration. We can solve for [H⁺] using the following formula: [H⁺] = 10^(-pH) So, for our case: [H⁺] = 10^(-10.50) Calculating this, we get: [H⁺] ≈ 3.16 × 10⁻¹¹ mol/L

Step 2: Find the hydroxide ion concentration

Now that we have the H⁺ concentration, we can find the OH⁻ concentration using the relationship between H⁺ and OH⁻ concentration: [H⁺] × [OH⁻] = Kw where [OH⁻] is the OH⁻ concentration, and Kw is the ion product of water, which is constant and equals 1 × 10⁻¹⁴ at 25°C. We can use this equation to solve for [OH⁻]: [OH⁻] = Kw / [H⁺] For our case: [OH⁻] = 1 × 10⁻¹⁴ / 3.16 × 10⁻¹¹ Calculating this, we get: [OH⁻] ≈ 3.17 × 10⁻⁴ mol/L

Step 3: Find the concentration of Sr(OH)₂

In an aqueous solution, strontium hydroxide (Sr(OH)₂) dissociates into its ions as follows: Sr(OH)₂ → Sr²⁺ + 2OH⁻ From the stoichiometry of the reaction, we know that one molecule of Sr(OH)₂ dissociates to produce two OH⁻ ions. Therefore, the concentration of Sr(OH)₂ is half the concentration of OH⁻: [Sr(OH)₂] = [OH⁻] / 2 For our case: [Sr(OH)₂] = 3.17 × 10⁻⁴ mol/L / 2 Calculating this, we get: [Sr(OH)₂] ≈ 1.59 × 10⁻⁴ mol/L So, the concentration of the aqueous Sr(OH)₂ solution with a pH of 10.50 is approximately 1.59 × 10⁻⁴ mol/L.

Key Concepts

Hydrogen Ion Concentration

When solving for the concentration of hydrogen ions in a solution, understanding how it relates to pH is essential. The pH of a solution is a measure of its acidity, and it is based on the concentration of hydrogen ions (\([H^+]\)). A lower pH indicates a higher concentration of hydrogen ions, meaning the solution is more acidic. Conversely, a higher pH means the solution is more basic, possessing fewer hydrogen ions.

To find the \([H^+]\) concentration from a given pH, we use the formula:

• \([H^+] = 10^{-pH}\)

This formula comes from the definition of pH: \( \text{pH} = -\log([H^+]) \) . The use of powers of 10 in the formula makes it easy to compute the \([H^+]\) concentration directly from the pH value. In our exercise, with a pH of 10.50, the \([H^+]\) is approximately \( 3.16 \times 10^{-11} \text{ mol/L} \). Calculating \([H^+]\) is a step towards determining other ion concentrations in the solution.

Hydroxide Ion Concentration

Understanding the concentration of hydroxide ions (\([OH^-]\)) is crucial in determining the basic nature of a solution. The relationship between hydrogen ions and hydroxide ions comes from the ion product constant of water (\(K_w\)), which at 25°C is \(1 \times 10^{-14} \text{ mol}^2/\text{L}^2\). This implies that the multiplication of \([H^+]\)and \([OH^-]\) in any aqueous solution at this temperature equals this constant. Thus, if one ion concentration is known, the other can be easily calculated.

For hydroxide ions, we have:

• \([OH^-] = \frac{K_w}{[H^+]} \)

Using the \([H^+]\) concentration from the earlier step (\( 3.16 \times 10^{-11} \text{ mol/L} \)), we can determine the \([OH^-]\) concentration to be approximately \( 3.17 \times 10^{-4} \text{ mol/L} \) . Knowing \([OH^-]\) concentration helps us understand the solution's alkalinity, and it is also vital for calculating the concentration of other components, like strontium hydroxide in our exercise.

Strontium Hydroxide Concentration

To find the concentration of strontium hydroxide (\( \text{Sr(OH)}_2\)) in a solution, we must consider its dissociation in water. Strontium hydroxide is a strong base and completely dissociates into strontium ions (\( \text{Sr}^{2+}\)) and hydroxide ions (\( \text{OH}^-\)) when dissolved. The chemical equation for this dissociation is:

• \( \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- \)

This indicates that each molecule of strontium hydroxide produces two hydroxide ions. As such, the concentration of \( \text{Sr(OH)}_2 \) is half the concentration of \( \text{OH}^- \). Knowing the \([OH^-]\) from the previous step (\( 3.17 \times 10^{-4} \text{ mol/L} \)), we can calculate:

• \([\text{Sr(OH)}_2] = \frac{[OH^-]}{2} \)

This gives us a concentration of about \( 1.59 \times 10^{-4} \text{ mol/L} \) for \( \text{Sr(OH)}_2 \). Understanding these relationships and dissociations helps in comprehensively solving such exercises.