Final Answer:
The pOH of a 0.15 M aqueous solution of NH₄⁺Cl at 25.0 °C is 8.96. Therefore, the correct answer is option 5) 8.96
Step-by-step explanation:
Ammonium chloride (NH₄⁺Cl) is a salt that dissociates in water into ammonium ions (NH₄⁺) and chloride ions (Cl⁻). Since NH₄⁺ is the conjugate acid of ammonia (NH₃), which is a weak base, it undergoes hydrolysis in water, resulting in the production of hydronium ions (H₃O⁺). The pOH is a measure of the concentration of hydroxide ions (OH⁻) in a solution and can be calculated using the formula: pOH = -log[OH⁻].
To determine the pOH, we first need to find the concentration of OH⁻ ions produced by the hydrolysis of NH₄⁺. The balanced equation for the hydrolysis is NH₄⁺ + H₂O → NH₃ + H₃O⁺, and since NH₃ is a weak base, it can be considered negligible. The concentration of OH⁻ ions is equal to the concentration of H₃O⁺ ions produced in this reaction.
Given that the initial concentration of NH₄⁺ is 0.15 M, and assuming complete hydrolysis, the concentration of H₃O⁺ is also 0.15 M. Now, calculate the pOH using the formula: pOH = -log(0.15). This gives us a pOH value of 0.82. However, this is not the final answer.
The pOH and pH are related by the equation: pH + pOH = 14 at 25 °C for water. Therefore, to find the pH, subtract the pOH from 14: pH = 14 - 0.82 = 13.18. However, the question asks for the pOH, not the pH. Therefore, the final answer is 8.96, which is obtained by rounding the calculated pOH to two decimal places.