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Answer :
Final answer:
The correct option isD. 10.7 The pH of a 2.5 x 10⁻⁴ M Sr(OH)₂ solution is 10.7, calculated by determining the OH⁻ concentration and using the pOH to find the pH.
Explanation:
To determine the pH of a 2.5 x 10⁻⁴ M Sr(OH)₂ solution, we must first understand that Sr(OH)₂ dissociates into one Sr²⁺ ion and two OH⁻ ions in water. This implies that the concentration of OH⁻ will be twice the concentration of the Sr(OH)₂, which results in an OH⁻ concentration of 5.0 x 10⁻⁴ M (2 x 2.5 x 10⁻⁴ M).The next step is to calculate the pOH of the solution: pOH = -log [OH⁻] = -log [5.0 x 10⁻⁴] = 3.3. Finally, we use the relationship pH + pOH = 14 to find the pH: pH = 14 - pOH = 14 - 3.3 = 10.7.Therefore, the pH of a 2.5 x 10⁻⁴ M Sr(OH)₂ solution is 10.7.
To find the pH of a 2.5 x 10-4 M Sr(OH)2 solution, we need to determine the concentration of OH- ions. Since the concentration of Sr2+ ions is negligible compared to OH- ions, we can assume that all of the Sr(OH)2 dissociates into OH- ions. Therefore, the concentration of OH- ions is equal to 2 times the concentration of Sr(OH)2. Hence, [OH-] = 2 x 2.5 x 10-4 M = 5.0 x 10-4 M.
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Answer:
The answer is D. 10.7
Explanation:
Determine the concentration of O H^ - ions in the solution by considering the stoichiometry of the dissociation reaction.
The concentration of O H^ - ions is 5 * 10 ^ - 4 * M
Calculate the pOH of the solution using the concentration of O H^ - ions.
The pOH of the solution is 3.30.
Use the relationship between pH and pOH to find the pH of the solution.
The pH of the solution is 10.70.
