Final answer:
The pH at the equivalence point for the titration of HNO₂ with KOH can be calculated by determining the concentration of the NO₂- ion, using the Ka value of HNO₂ to find the concentration of OH- ions, and then calculating the pOH and pH of the solution, which will be basic and greater than 7.
Step-by-step explanation:
To determine the pH at the equivalence point for the titration of HNO₂ (nitrous acid) with KOH (potassium hydroxide), we need to consider that at the equivalence point, all of the HNO₂ has reacted with KOH to form water (H₂O) and a salt, KNO₂. Since HNO₂ is a weak acid and KOH is a strong base, the resulting solution will be slightly basic. The pH can be found by calculating the concentration of the A- ion (NO₂- in this case), and then using the Ka value of HNO₂ to find the concentration of OH- ions, from which we can calculate the pOH, and subsequently the pH. First, calculate the moles of HNO₂ and KOH: Moles of HNO₂ = 0.0380 L × 0.188 M = 0.007144 mol Moles of KOH = 0.007144 mol (since it's a 1:1 reaction).
Then, calculate the concentration of NO₂- in the solution at the equivalence point: Volume of the solution at equivalence = 38.0 mL of acid + volume of KOH added Assume the volume of KOH added is x mL: Concentration of NO₂- = 0.007144 mol / (0.038 L + x L) The concentration of NO₂- will be equal to the concentration of OH- at the equivalence point because the salt hydrolyzes in water.
We can then use the Ka for HNO₂ and the stoichiometry of the reaction to find the pOH and consequently the pH: Ka × Kb (equilibrium constant for the reaction of NO₂- with water) = × × (Kw/Ka), where Kw is the ion product of water. Kw = 1.00 × 10⁻¹⁴ at 25°C. Finally, using the values of Ka and the calculated concentration of NO₂-, we can solve for the concentration of OH- and find the pOH and pH from there. pOH = -log[OH-], and pH = 14 - pOH. The pH will be greater than 7, indicating a basic solution.