Calculate the pH of a solution prepared by dissolving 0.750 mol of NH₃and 0.250 mol of NH₄Cl in water sufficient to yield 1.00 L of solution. The K_bof ammonia is 1.77 × 10^-⁴. A) 8.78 B) 0.89 C) 9.73 D) 4.27 E) 5.22

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First, we need to find the concentration of NH3 produced from the reaction of NH₃and water. NH₃+ H2O → NH3 + H_- Since we are starting with 0.750 mol of NH₃, the concentration of NH3 is also 0.750 mol/L. Next, we need to find the concentration of NH4+ produced from the reaction of NH₄Cl and water. NH₄Cl + H2O → NH4+ + Cl- Since we are starting with 0.250 mol of NH₄Cl, the concentration of NH4+ is also 0.250 mol/L. Since NH3 is a weak base and NH4+ is its conjugate acid, we need to use the expression for the base dissociation constant, Kb, to find the concentration of OH- in the solution. Kb = Kw/Ka = (1.0 x 10^-14)/(1.77 x 10^-5) = 5.65 x 10^-10 Kb = [NH4+][OH-]/[NH3] We can assume that the amount of NH4+ that ionizes to form OH- is negligible compared to the amount of NH3, so we can simplify the expression to Kb ≈ [OH-]^2/[NH3] Solving for [OH-], we get [OH-] = √(Kb x [NH3]) = √(5.65 x 10^-10 x 0.750) = 1.09 x 10^-5 mol/L Finally, we can find the pH of the solution using the expression: pH = 14 - pOH = 14 - (-log[OH-]) = 14 - (-log(1.09 x 10^-5)) = 9.73 Therefore, the pH of the solution is 9.73, and the answer is C.

Calculate the PH of a Solution Prepared by Dissolving 0