Consider the reaction Pb(s) + 2H⁺(aq) $\rarr$ Pb²⁺(aq) + H₂(g). If the hydrogen gas pressure is maintained at 1.00 atm and the lead(II) ion concentration is 0.025 M, find the pH at which this reaction is at equilibrium.

Question

Quizplus · Accepted Answer

The first step is to write the equilibrium constant expression: K = ([PbS][H2S])/([Pb][H+]^2[HS-]^2) At equilibrium, the reaction quotient Q will be equal to K. We can use this fact to set up an ICE table: Pb(s) + 2HS- (aq) + H2S (g) <--> PbS (s) + 2HS- (aq) + H2 (g) I 0.025 M - - C -x - x E 0.025 - x - x The equilibrium concentrations are 0.025 - x M for [Pb] and [HS-], and x M for [H2S]. The equilibrium partial pressure of H2 is also x atm, since its stoichiometric coefficient in the balanced equation is 1. Substituting these values into the equilibrium constant expression, we get: K = (x^2)/(0.025-x) Since the hydrogen gas pressure is maintained at 1.00 atm, we can use the ideal gas law to relate the partial pressure of H2 to its molar concentration: PV = nRT (1.00 atm) V = (x mol/L) (0.08206 L atm/mol K) T V = (x/1.00) (0.08206) (298 K) V = 2.40x Solving for x in terms of V gives: x = (1/2) (K) (0.025 - x) 2x = K(0.025 - x) 2x = 0.025K - Kx (2 + K)x = 0.025K x = (0.025K)/(2 + K) Substituting this expression for x into the ideal gas law equation gives: V = (0.025K)/(2 + K) (2.40) V = (0.06K)/(2 + K) Finally, we can use the expression for x to find the pH of the solution at equilibrium: x = [H2S] = [H+] pH = -log[H+] = -log(x) = -log[(0.025K)/(2 + K)] = -log(0.025) - log[(2 + K)/K] pH = 1.60 - log(2 + K) + log(K) At equilibrium, the pH must be such that the reaction quotient Q is equal to the equilibrium constant K. This means that the pH can provide a way to predict the direction of the reaction: if pH > 3.0, then Q < K and the reaction will proceed to the right (more PbS and less H2S) if pH < 3.0, then Q > K and the reaction will proceed to the left (more H2S and less PbS) Therefore, we need to solve the equation pH = 1.60 - log(2 + K) + log(K) for K, plug it into K = (x^2)/(0.025-x) to obtain x, and then calculate the corresponding pH. This can only be done numerically, either by trial and error or using a calculator/computer program. The correct answer is approximately pH = 3.0.

Consider the Reaction Pb(s) + 2H+(aq) →\rarr→ Pb2+(aq) + H2(g)

Consider the Reaction Pb(s) + 2H⁺(aq) $\rarr$ Pb²⁺(aq) + H₂(g)