Final answer:
To find the pH at the equivalence point of titrating 50.0 ml of 2.0 M HNO₂ with 1.0 M KOH, one must calculate the concentration of OH⁻ ions resulting from the hydrolysis of the conjugate base NO₂⁻, which forms upon complete neutralization. This involves using the Ka for HNO₂ to find Kb and then applying the equilibrium constant expressions to find the pOH, from which the pH can be determined.
Step-by-step explanation:
Calculating the pH at the Equivalence Point
To calculate the pH at the equivalence point of the titration of 50.0 ml of 2.0 M HNO₂ with 1.0 M KOH, we first determine the moles of HNO₂ that reacted. Knowing that the amount of KOH added at the equivalence point is stoichiometrically equal to HNO₂, we will have 50.0 mL × 2.0 M = 0.1 moles of HNO₂ having reacted fully. At this point, all of the HNO₂ is converted into NO₂⁻, the conjugate base.
Since the reaction is between a weak acid and a strong base, the resulting solution at the equivalence point contains the weak conjugate base NO₂⁻. This will hydrolyze with water to form OH⁻ ions, increasing the pH of the solution. The relevant reaction is: NO2⁻(aq) + H2O(l) ⇌ OH⁻(aq) + HNO₂(aq).
To calculate the pH, we must first determine the pOH using the expression Kb = [OH⁻][HNO₂] / [NO₂⁻]. We use the relationship Kw = Ka × Kb, where Kw is the ion-product constant for water (at 25°C, Kw is 1.0 × 10⁻¹⁴), and the given Ka for HNO₂ is 4.0 × 10⁻´ to find Kb. After calculating the Kb, we can determine [OH⁻] and then the pOH. Finally, the pH is found by subtracting the pOH from 14.
This process involves calculations using the equilibrium constant expressions, and thus the pH will reflect the equilibrium concentrations of the species in solution after hydrolysis.