Final Answer:
The pH after adding 7.50 ml of 0.200 M HCl to a 20.00 ml sample of 0.150 M NH₃ is 9.74.
Step-by-step explanation:
To find the pH after adding HCl to NH₃, we need to determine the moles of NH₃ initially present and the moles of HCl added. Initially, moles of NH₃ = Molarity × Volume = 0.150 mol/L × 0.020 L = 0.003 moles. The moles of HCl added = Molarity × Volume = 0.200 mol/L × 0.0075 L = 0.0015 moles. NH₃ reacts with HCl in a 1:1 ratio, so after the reaction, 0.0015 moles of NH₃ are neutralized, leaving 0.003 - 0.0015 = 0.0015 moles of NH₃ in excess.
To find the concentration of NH₃ after the reaction, divide the remaining moles by the total volume. Total volume after HCl addition = 20.00 ml + 7.50 ml = 27.50 ml = 0.0275 L. Concentration of NH₃ = Moles/Volume = 0.0015 moles / 0.0275 L = 0.0545 M.
Now, to find the pOH, use the Kb of NH₃ to find the OH⁻ concentration. Kb = [NH₄⁺][OH⁻] / [NH₃]. At equilibrium, [NH₄⁺] ≈ [OH⁻] because NH₄⁺ is from the conjugate acid of NH₃. Thus, Kb = [OH⁻]² / [NH₃]. Rearranging the equation gives [OH⁻] = √(Kb × [NH₃]) = √(1.8 × 10⁻⁵ × 0.0545) = 0.00416 M.
Finally, to find the pH, use pH + pOH = 14. pH = 14 - pOH = 14 - (-log[OH⁻]) = 14 - (-log 0.00416) ≈ 9.74.
This pH value (9.74) indicates that the solution is basic due to the presence of excess NH₃, making it more alkaline after the addition of HCl.