Question

Suppose you have 500. mL of a solution of 0.200 M sodium benzoate (C6H5CO2Na), to which you add 100. mL of a 0.400 M HCl solution. After the resulting mixture has reached equilibrium, what is the pH of this solution? Ka of benzoic acid (C6H5CO2H) = 6.3 x 10−5 at 25°C.

Suppose you have 500. mL of a solution of 0.200 M sodium benzoate (C6H5CO2Na), to which you add 100. mL of a 0.400 M HCl solution. After the resulting mixture has reached equilibrium, you add an additional 50.0 mL of the 0.400 M HCl solution into the solution and allow it to re-establish equilibrium. What is the pH of the solution now? Ka of benzoic acid (C6H5CO2H) = 6.3 x 10−5 at 25°C. (1) 2.98 (2) 3.57 (3) 3.80 (4) 4.02 (5) 4.90

Answer 1