Question

4. (14 pts) Assume that you have a 0.250 L buffer solution that is 0.40 M acetic acid, HC₂H₃O₂, and 0.50 M acetate ion,

HC2H3O2, and 0.50 M acetate ion, C2H30. Calculate the pH after 100 mL of 0.10 M HCl is added to this buffer. HC,H,02(aq) + H,0(I) C,H30, (aq)+H,0. (aq) K.-1.8x105

Answer 1

Ans. As shown in the table-

            Equilibrium [AH] = 0.3143 M

            Equilibrium [A^-] = 0.3286 M

Conjugate base A CH3COO pKa =-log Ka = 4.74 Initial [A-]= 0.500 M; [HCl]= Reaction: AHCI > AH CI Weak acid AH CH3COOH Ka of acid AH = 1.800E-05 Given: Initial AH- kb = 10-14 / Ka = 5.556E-10 0.400 M 0.10 M Vol of original buffer= 250.00 mL = 0.250 L Vol. of HCl added = 0.00 ml Total vol.= HCI moles= 0.000 L 0.250 L 0.000 mol HCl moles= 100.00 mL0.100 L Total vol.- 0.350 L 0.010 mol Initial Moles of AH Initial Moles of A- Mol of A consumed Mol of AH Formed Total moles of AH = (initial + Formed) Total moles of A-= (Initial-Consumed) Remaining moles of NaOH = In-Con. = 0.1000 mol 0.1250 mol 0.0100 mol 0.0100 mol 0.1100 mol 0.1150 mol 0.00 mol 0.1000 mol 0.1250 mol 0.0000 mol 0.0000 mol 0.1000 mol 0.1250 mol 0.0000 mol 0.4000 M 0.5000 M In mixture, Final [AH] = In mixture, Final A-] = In mixture, final [HCl] = 0.3143 M 0.3286 M 0.000 M 0.000 M

Now, calculate pH using HH equation as follow-

Henderson- Hasselbalch equation for Base: pH pKa + log (A1/[AHI)Equation 1 Where, [AHWeak acid CH3COOH AConjugate base - CH3COO , Ka- Ka x Kb-10-14 1014 / Kb Kb5.56E-10 : Kb 1014/ KapOH 14.00 So, pKalog Ka4.745 Given, Ka 1.80E-05 AH 0.3143 M A pH of Resultant buffer: Putting the values in equation1 pHH4.745og 0.3286 M 0.329 ' 4.745 0.0193 4.764 0.314