Question

A buffer of pH 3.270 is prepared by adding 85.00mL of 0.1120M HA with 50.00mL of 0.2155M NaOH. What is the pKa of the acid, HA? How would it be solved if there is excess in strong base or acid?

Answer 1

apply buffer equation

mmol of HA = MV = 85*0.112 = 9.52

mmol of base = MV = 0.2155*50 = 10.775

note that,

NaOH, base > HA

this is NOT a buffer, this will occur:

- Addition of Some NaOH --> the solution becomes a buffer

-Exact addition of NaOH --> this is neutralization

- Extra addition of NaOH --> the solution becomes basic, due to OH-

therefore, the problem is incorrect, it is impossible to have a pH = 3.27 with such amounts of acid/base.

for pKa:

pH = pKa + log(A- / HA)

since there is no HA after NaOH adidtion, the buffer equaztion cant be applied.

Also, if this were a strong acid/base problem,

simply calculate the amount of H+ and OH- left after reaction

then add total volume, clacualte concentration, ang get pH