15.0 mL of 0.50 M HCl is added to a 100.0-mL sample of 0.484 M HNO₂ (K_a for HNO₂ = 4.0 × 10^-4).What is the equilibrium concentration of NO₂^- ions?

Question

Quizplus · Accepted Answer

The correct answer is D because the equilibrium concentration of NO2- ions is 2.6 × 10-3 M. This can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In this case, the pKa of HNO2 is 4.0 × 10-4, the concentration of HCl is 0.50 M, and the concentration of HNO2 is 0.484 M. Plugging these values into the equation, we get: pH = 4.0 × 10-4 + log([NO2-]/[HNO2]) = 3.68. Solving for [NO2-], we get: [NO2-] = [HNO2] × 10^(pH - pKa) = 0.484 M × 10^(3.68 - 4.0 × 10-4) = 2.6 × 10-3 M.

150 ML of 0