What is the reactant concentration, [HA], in HA ⇌ H⁺ + A^- , when [H⁺] = 0.010 M and K = 2.05?

A) 1.00 × 10^-4
B) 4.87 × 10^-5
C) 4.87 × 10^-3
D) 1.00 × 10^-3

Given the equilibrium constant (K) is 2.05 and the concentration of the product [HS^+ + AS^-] is 0.010 M, we can set up the equilibrium expression for the dissociation of HA as K = [HS^+][AS^-] / [HA]. Assuming complete dissociation of HA into HS^+ and AS^-, both HS^+ and AS^- would have concentrations of 0.010 M. Thus, K = (0.010)(0.010) / [HA] = 2.05. Solving for [HA] gives [HA] = (0.010)(0.010) / 2.05 = 4.87 × 10^-5 M.