Determine the pOH of a 0.188 mol L^-1 NH₃ solution at 25 °C. The K_bof NH₃ is 1.76 × 10^-5. A) 5.480 B) 2.740 C) 8.520 D) 11.260 E) 12.656

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Quizplus · Accepted Answer

Step 1: Write the balanced chemical equation for the dissociation of NH₃ in water: NH₃ + H2O → H3O+ + ₃- Step 2: Write the expression for K_band solve for [OH-]: K_b= [H3O+][₃-]/[NH₃] 1.76 × 10^-5= (x)(x)/(0.188-x) x = 3.51 × 10^-3 mol/L Step 3: Calculate pOH and pH: pOH = -log[OH-] = -log(3.51 × 10^-3) = 2.740 pH = 14 - pOH = 11.260 Therefore, the correct answer is B) 2.740.

Determine the POH of a 0