Calculate the pOH of a solution that contains 3.9 × 10^-4 mol L^-1 H₃O⁺ at 25 °C. A) 4.59 B) 3.31 C) 10.59 D) 9.14 E) 0.59

Question

Quizplus · Accepted Answer

To find the pOH of the solution, we need to use the equation pOH = -log[OH-]. 
Firstly, we need to calculate the concentration of OH- ions in the solution using the given concentration of H3O+. The equation for the autoionization of water is: 
H2O + H2O ⇌ H3O+ + OH- 
Kw = [H3O+][OH-] = 1.0 × 10^-14 
At 25 °C, Kw = 1.0 × 10^-14. 
Therefore, [OH-] = Kw/[H3O+] = 1.0 × 10^-14/3.9 × 10^-14 = 0.256 M 
Now, we can calculate the pOH of the solution: 
pOH = -log[OH-] = -log(0.256) = 0.59 
Finally, we need to subtract the pOH value from 14 to find the pH of the solution: 
pH + pOH = 14 
pH = 14 - pOH = 14 - 0.59 = 13.41 
However, the question asks for the pOH value, not pH. So, the best choice is C, which is the value we just calculated for pOH.

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