Question

Calculate the volume of the 0.1305 M NaOH solution that must be added to the 35 00 mL of 0 1212 M acetic acid (CH COOH) solution so that the pH of the solution is equal to 4 744 Also calculate the mole fraction of CH,COOH at that volume (8+6)

Answer 1

pKa of acetic acid = 4.744

pH of solution given = 4.744

pH is exaclty equal to pKa. so it is half equivalence point

here mmoles of acid left = mmoles salt formed

mmoles of acid CH3COOH = 35 x 0.1212 = 4.242

half equivalence point = 4.242 / 2 = 2.121

2.121 millimoles must be equal to NaOH

2.121 = M x volume

2.121 = 0.1305 x volume

volume = 16.25 mL

volume of NaOH needed = 16.25 mL

now part two

millimoles of acid = 2.121

mimmoles of NaOH also = 2.121 at this volume

mole fraction of acid = moles of acid / total moles

                                 = 2.121 / 2.121 + 2.121

                                = 1/2

                                = 0.5

mole fraction of CH3COOH = 0.50