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Chapter 13: Problem 94

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

Short Answer

Expert verified
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 by step solution

01

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
02

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
03

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.

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