Imidazole (C3H4N2, MW = 68.07816, Kb = 9.80 x 10–8) can be used to create buffers at physiological pH’s. How many grams of imidazole and how many milliliters of 6.00 M HCl would be required to make 500.0 mL of 0.1000 M imidazole buffer with a pH of 7.20? The image depicts a chemical equilibrium involving imidazole. On the left, the structure of imidazole is shown as a five-membered aromatic ring containing two nitrogen atoms—one with an NH group and the other as part of the ring. This structure is labeled "Imidazole." On the right is the protonated form of imidazole, where one of the nitrogen atoms carries an extra hydrogen, resulting in a positive charge (NH⁺). The equilibrium between these two forms is indicated by a double-headed arrow labeled "K_b," which stands for the basicity constant, representing the equilibrium constant for the deprotonation of the imidazole. This diagram highlights the acid-base chemistry of imidazole in equilibrium with its conjugate acid.

Buffer are solution which resist change in PH when acid Or base are added Mass of imidazole =x gm Molar mass =68.07816 g/mol Moles of imidazole = Mass taken/ Molar mass Moles of imidazole =x/68.07816 Moles of imidazole =0.01469 x molMolarity Of HCl=6M Volume of HCl taken= V litre So moles of HCl=Molarity *Volume Moles of HCl= 6 mol/litre*V litre =6V mol As HCl react with imidazole in1:1 ratio So moles of imidazole conjugate base =6V So 0.01469 x= 0.0308 X= 0.0308/0.01469 gm X=2.096 gm And 6V=0.0191 v= 0.0191/6=3.183*10-3 L=3.183 mlSo we require 3.183 ml of HCl and 2.096 gm of imidazole to prepare the buffer