Question

A) In proteins often a gly amino acid can be successfully substitute for a val amino acid but val can rarely successfully substitute for a gly. Why do you think this is true?

B)Describe how you would make 2 L of a 0.1 M glycine buffer, pH 9.0, using glycine (molecular weight in the zwitterionic form is 75.07 g/mol) and 1.0 M NaOH. For glycine the pKa is 9.6.

Answer 1

we know, From Henderson equation,

pH=pKa +log([salt]/[Acid]);

9 =9.6 + log(S/A); which gives ration of concentration of base by acid i.e, S/A= 0.25; or S = 0.25 x A; A = 4S

Now, moles of buffer, 0.2 L x 0.1 mol/L = 0.02 moles

so we can write, 0.02 mol = S+A; or 0.02 mol = S+4S = 5S

or, S = 4 x 10^-3 mol, so A = 1.6 x 10^-2 mol

so A or glycine required = 1.6 x 10^-2 mol + 4 x 10^-3 mol = 20 x 10^-3 mol = 20 x 10^-3 mol x 75.07 = 1.5 g

mol of salt will be the mol of NaOH;

so NaOH required = 4 x 10^-3 mol x 40 = 0.16 g

so mix these quantities in 2 L to make the requierd buffer.

34.37g = 119.96 (mol wt of acid i.e, NaH2PO4) x 0.286

12.57g = 141.96 (mol wt of base or Na2HPO4) x 0.885 (Base)

that is mix the above quantities of the two components in the 750 mL solvent to obtain the 0.5 M of phosphate buffer.

do the same thing, just by putting pH 7.8 in Henderson equation for the second one.

plz let me know if you have any confusion.