A 100.0 mL sample of 0.10 mol L^-1 NH₃ is titrated with 0.10 mol L^-1 HNO₃. Determine the pH of the solution after the addition of 200.0 mL of HNO₃. The K_b of NH₃ is 1.8 × 10^-5. A) 6.44 B) 1.48 C) 2.00 D) 12.52 E) 12.00

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First, let's write out the balanced chemical equation for the reaction that occurs during the titration: L^-1 NH₃ + L^-1 HNO₃ → L^-1 NO₃ + L^-1 H2O At the beginning of the titration, we have a solution of NH₃, which is a weak base, and we add a strong acid, HNO₃. As the acid is added, it will react with the base to form the conjugate acid of the base, NH₃H, and water. NH₃H is a weak acid, so it will partially dissociate in water to form H+ and the conjugate base, NH₃. We can use the equilibrium constant expression for this acid dissociation reaction to calculate the pH of the solution after the addition of the acid: K_b = [H+][NH₃] / [NH₃H] We can assume that the concentration of NH₃H at the beginning of the titration is negligible, since we are titrating a weak base with a strong acid. This means that the initial concentration of NH₃ is 0.10 mol/L. After the addition of 200.0 mL of the 0.10 mol/L acid, the total volume of the solution is 100.0 mL + 200.0 mL = 300.0 mL. The amount of acid added is: (0.10 mol/L) x (0.200 L) = 0.020 mol This means that the amount of base reacted is also 0.020 mol, and the final concentration of NH₃ is: (0.10 mol - 0.020 mol) / 0.300 L = 0.267 mol/L Now we can substitute the concentrations into the equilibrium constant expression and solve for [H+]: 1.8 × 10^-5 = [H+][0.267 mol/L] / [0.10 mol/L - 0.020 mol/L] [H+] = 1.48 x 10^-6 M Finally, we can take the negative logarithm of [H+] to calculate the pH: pH = - log[H+] = 1.48 + 6 = 7.48 Since pH is defined as -log[H+], we need to subtract the exponent from 14 to get the pH value. Therefore, the pH of the solution after the addition of the acid is: pH = 14 - 7.48 = 6.52 The closest answer choice to this pH value is B (1.48), which is the correct answer.

A 1000 ML Sample of 0