Question

A chemist titrates 90.0 mL of a 0.5705 M carbonic acid (H,CO,) solution with 0.6274 M KOH solution at 25 °C. Calculate the pH at equivalence. The pk of carbonic acid is 3.60. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. DH-

Answer 1

Given data : Carbonic acid HaCo3 = 0.5705M Chemist titrcut le =90.0 ml KOH 0.627M Carbonicacid = 3.60 H₂CO3 = 90 ml of 0.5705 m noic moles Hace =0.5705 x 0.90 L = 0.5135 molesi Kolt = 0.6274M pka = 3.60 At @wwilance point : 0.5 705 X 90 . volum of kou = 0.6974 51.345 016274 81.837m ! 81.eu mi

Total volume = 90 + 81.84 =171.80ml Io.AL Concentration - C = number of moles volume 0.5135 0:171 = 3.00 M. AL Ewilance point: pH = 7+1 (pka + loge pit - 2+ 3.60 +log (3.00) . 2- 4.0+H12 = 2.03856 +7