What is the pH of a solution that consists of 0.50 M H₂C₆H₆O₆ (ascorbic acid) and 0.75 M NaHC₆H₆O₆ (sodium ascorbate)? For ascorbic acid, K_a = 6.8 × 10^-5 A) 3.76 B) 3.99 C) 4.34 D) 4.57 E) 5.66

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First, we can write the dissociation equation for ascorbic acid: H₂C₆H₆O₆ ⇌ H+ + ₂C₆- + ₆O2- The equilibrium constant, Ka, is given as 6.8 × 10^-5. Next, we can write the expression for Ka: Ka = [H+][₂C₆-][₆O2-] / [H₂C₆H₆O₆] We know the concentrations of both ascorbic acid and sodium ascorbate, so we can use an ICE table (Initial, Change, Equilibrium) to solve for the concentration of each species at equilibrium: Initial: [H₂C₆H₆O₆] = 0.50 M (ascorbic acid) [₆H₆O₆] = 0.75 M (sodium ascorbate) [H+] = 0 (since the solution is not acidic initially) [₂C₆-] = [₆O2-] = 0 (since they are not present initially) Change: - x (since the concentration of H+ goes down as ascorbic acid dissociates) + x (since the concentration of H₂C₆- increases as ascorbic acid dissociates) + x (since the concentration of ₆O2- increases as ascorbic acid dissociates) Equilibrium: [H₂C₆H₆O₆] = 0.50 - x [₆H₆O₆] = 0.75 + x [H+] = x [₂C₆-] = x [₆O2-] = x We can substitute these equilibrium concentrations into the expression for Ka: Ka = x * x / (0.50 - x) Since we know the value of Ka (6.8 × 10^-5) and the concentration of ascorbic acid (0.50 M), we can solve for x: 6.8 × 10^-5 = xS1U1P20S1S1S1S1S1S1S1S1S1S1P2 / (0.50 - x) x = 1.78 × 10 M Finally, we can use the concentration of H+ to calculate pH: pH = -log[H+] = -log(1.78 × 10) = 4.34 Therefore, the answer is C.

What Is the PH of a Solution That Consists of 0.50