Calculate the pH of a solution prepared by dissolving 0.270 mol of formic acid (HCO₂H)and 0.230 mol of sodium formate (NaCO₂H)in water sufficient to yield 1.00 L of solution. The K_a of formic acid is 1.77 × 10^-4. A)2.099 B)10.318 C)3.682 D)2.307 E)3.952

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First we need to write the equation for the dissociation of formic acid: HCO₂H ⇌ H+ + CO₂¯ We can then use the equilibrium expression: K_a = [H+][CO₂¯] / [HCO₂H] Since we are given the amounts of formic acid and sodium formate, we can calculate the initial concentrations of each species: [HCO₂H] = 0.270 mol / 1.00 L = 0.270 M [CO₂¯] = 0.230 mol / 1.00 L = 0.230 M We can assume that the concentration of H+ is initially zero, since the solution is not acidic before any dissociation occurs. Therefore, we can simplify the equilibrium expression to: K_a = x^2 / (0.270 - x) where x is the concentration of H+ and (0.270 - x) is the concentration of HCO₂H and CO₂¯ (assuming that all of the sodium formate dissociates to form the formate ion). We can solve for x using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a where a = 1, b = 0, and c = -K_a(0.270). Plugging in these values, we get: x = 0.706 M Finally, we can calculate the pH: pH = -log[H+] pH = -log(0.706) pH = 3.682 Therefore, the answer is choice C.

Calculate the PH of a Solution Prepared by Dissolving 0