Question

How much 5 M KOH must be added to 1 L of 0.1M histidine at pH 6 to bring it's pH exactly to 6.5?

Answer 1

Meet the actors. Glycine is aminoacetic acid (NH2-CH2-COOH). If we start with 0.1 M glycine in pure water, the equilibrium will be
[H+][GlyO-]/[GlyOH]= 2.5x10^-10
So little of the 0.1 M glycine will dissociate, that we can write
[H+]^2/[0.1]= 2.5x10^-10 and [H+]=1.6x10^-6 mole/L.

So even to get to pH 6, we must add a base like KOH. It reacts with the acid to form GlyO- ion on a 1 to 1 basis with the acid lost. To compute this situation at pH 6, we can write
[1x10^-6][GlyO-]/[0.1-GlyO-] =2.5x10^-10
[GlyO-]=0.02 mole/L and [GlyOH]=0.08 mole/L
To get to pH 6.5, we can repeat this calc as:
[1x10^-6.5][GlyO-]/[0.1-GlyO]=2.5x10^-10
[GlyO-]=0.071mole/L and [GlyOH]=0.029 mole/L appx.
So in a 1 L solution, we have had to use 0.051 moles of KOH. Since moles= volume x molarity for a unibasic cpd,
0.051 = Volume x 5, and volume=6.5 mL.

Note that 6.5 mL can be neglected compared with 1 L; otherwise the conc would have to be adjusted for dilution