What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH₃ with 5.00 mL of 0.10 M NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃. A)8.25 B)10.06 C)10.25 D)11.13

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First, we need to calculate the moles of each solution: Moles of NH₃ = (0.10 mol/L) x (50.00 mL/1000 mL) = 0.005 mol Moles of NH₄Cl = (0.10 mol/L) x (5.00 mL/1000 mL) = 0.0005 mol Since the volumes are additive, the total volume is 50.00 mL + 5.00 mL = 55.00 mL. Next, we need to calculate the total concentration of the solution, which is: Total moles = 0.005 mol + 0.0005 mol = 0.0055 mol Total volume = 55.00 mL = 0.055 L Total concentration = 0.0055 mol / 0.055 L = 0.1 mol/L Now we can use the expression for the ionization of NH₃: NH₃ + H2O ⇌ H3O+ + _b K_b = [H3O+][_b] / [NH₃] Assuming that the initial concentration of _b is negligible compared to the concentration produced by the ionization of NH₃, we can simplify the expression to: K_b = [H3O+]^2 / [NH₃] Solving for [H3O+], we get: [H3O+] = sqrt(K_b[NH₃]) [H3O+] = sqrt((1.8 x 10^-5)(0.1)) = 2.68 x 10^-3 mol/L Finally, we can calculate the pH: pH = -log[H3O+] = -log(2.68 x 10^-3) = 2.57 + 10 = 12.57 However, we need to remember that the ionization of NH₃ produces OH- ions as well as H3O+ ions: NH₃ + H2O ⇌ H3O+ + _b NH₃ + H2O ⇌ OH- + _b The concentration of OH- can be calculated using the expression for KW: KW = [H3O+][OH-] OH- = KW / [H3O+] = 1.0 x 10^-14 / 2.68 x 10^-3 = 3.73 x 10^-12 mol/L Taking the negative log of this concentration gives: pOH = -log[OH-] = -log(3.73 x 10^-12) = 11.43 Finally, we can use the expression: pH + pOH = 14 to calculate the pH: pH = 14 - pOH = 14 - 11.43 = 2.57 + 10 = 12.57 Therefore, the pH of the solution is 10.25 (Choice C).

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