Correct pH-pK, =-log(6.4 × 10-5) = 4194 = 4.19 s [HBs] = [Br]. where HBz\" C6H,C
ID: 996526 • Letter: C
Question
Correct pH-pK, =-log(6.4 × 10-5) = 4194 = 4.19 s [HBs] = [Br]. where HBz" C6H,CO,H and [Bz-] = C,HsCO2 Calculate the pH after 800% by moles) of the benzo c acid is converted to benzoate anson by add ton of a strong base Use he dissociation equilibrium to calculate the p Incorrect [Ba ] will ncrease to 0.180 Mand HBal will decrease to 0.0200 Mafter OH reacts completely with HBs. The Henderson-Hasselbalch + log (0.180) l equation is derived from pH-pK, +10g , pH-4.194 + log (0.0200) 5.148 = 5.15; assumptions good G Do the same as n part b, but use the following equilibrum to calculate the pH CaHs CO, (ag) +H20() CH.CO,Hag) +OH (a) pHExplanation / Answer
For the above ques the conc of benzoic acid being used need to be known, since that is not provided i will show an example of a similar prob to clear your doubt.
Calculate the pH after 10.0% (by moles) of the benzoic acid is converted to benzoate anion by addition of base. solution that is 0.136 M in C6H5CO2H (benzoic acid, Ka= 6.4 10-5) and 0.498 M in C6H5CO2Na.
Using the following dissociation equilibrium.
C6H5CO2H(aq) ---> C6H5CO2(aq) + H+(aq)
0.136 - 0.0136 --> 0.498 + 0.0136
0.1224 --> 0.5116
K = [H] [A] / [HA]
6.4 e-5 = [H] [A] / [HA]
6.4 e-5 = [H] [0.5116] / [0.1224]
H+ = 1.5312 e-5
pH = 4.81497