Final answer:
To determine the concentration of nitrite ion (NO₂¯) in the solution, we can use the relationship between the concentrations of nitrous acid (HNO₂) and nitrite ion and the equilibrium constant (K₂). The concentration of nitrite ion (NO₂¯) in the solution is approximately 6.562 x 10^(-4) M.
Step-by-step explanation:
To determine the concentration of nitrite ion (NO₂¯) in the solution, we can use the relationship between the concentrations of nitrous acid (HNO₂) and nitrite ion and the equilibrium constant (K₂) for the reaction: HNO₂ (aq) + H₂O(1) → H₃O⁺ (aq) + NO₂¯ (aq) Given that the solution has a pH of 3.87, we can calculate the concentration of hydronium ion (H₃O⁺) using the formula pH = -log[H₃O⁺]. Substituting the value pH = 3.87 into the formula gives us [H₃O⁺] = 10^(-pH), which is approximately 1.38 x 10^(-4) M. Using the equilibrium equation, we can set up an ICE table to determine the concentrations of HNO₂ and NO₂¯ at equilibrium.
The initial concentration of HNO₂ is given as 0.055 M, and since HNO₂ is a weak acid, we can assume that x moles of HNO₂ ionize to form x moles of H₃O⁺ and NO₂¯. Therefore, the equilibrium concentration of HNO₂ is 0.055 - x M, and the equilibrium concentrations of H₃O⁺ and NO₂¯ are both x M. Using the equilibrium constant expression K₂ = [H₃O⁺][NO₂¯] / [HNO₂], we can substitute the values from the ICE table and solve for x. Plugging in the values gives us K₂ = (x)(x) / (0.055 - x), and rearranging the equation gives us 4.5 x 10^(-4) = x^2 / (0.055 - x).
Since x is small compared to 0.055, we can assume that 0.055 - x is essentially 0.055. Simplifying the equation further and solving for x using quadratic equation methods gives us x ≈ 6.562 x 10^(-4) M. Therefore, the concentration of nitrite ion (NO₂¯) in the solution is approximately 6.562 x 10^(-4) M.