[Chemistry] Calculate the pOH value of the solution?

You have the anion (=base) and the zwitterion (=acid). The fact that the two form a column/base pair is recognized by the fact that their structures are only an H+ and thus are the equilibrium.

RCH(N+H3)CO2 ̄ H+ + RCH(NH2)CO2 ̄

which is described by the pK2.

So it is obvious to apply the Henderson–Hasselbalch equation:

pH = pK2 + lg(base/acid) = pK2 + lg(anion/cyterion) = 9.1+lg(0.5/0.005) = 11.1

This would also be the right solution if the specified concentration (0.5 or 0.005 mol/l) Balanceconcentration of the two ions. However, it is probably meant that the two substances in these concentrations are in stock and then the equilibrium is established. Often, this difference can be ignored, but here it plays a certain role, because according to the above-calculated result c(OH ̄)=10 ̄pOH=0.0013 mol/l and therefore not substantially smaller than the concentration of the buffer substance itself (Zwitterion, 0.005 mol/l). It must therefore be expected that the result is inaccurate.

How to calculate it more accurately? Improving the HH equation is the Charlot equation that can be written in several forms; I prefer the following:

ctot=0.505 mol/l denotes the combined concentrations of the buffer species and c_B=0.5 mol/l of the basic form. The result is then pH=11.02, so it’s just a small correction that you might be able to ignore.

Furthermore, we neglected the pK1, but this is unproblematic because it is so much smaller than the pK2. At a pH around 11, the first dissociation stage (the equilibrium between cation and twitterion) is dissociated as well as completely; c(cation)=5⋅10 ̄12 mol/l, c(cyterion)=0.006 mol/l and c(anion)=0.499 mol/l in equilibrium.

If you want to annoy your teacher, then you will tell him the right result pOH= 3.0 if you want to behave with conformity then you will also reach 2.9 — Your education.