A buffer is made at pH 7.3. 0.15 M using imidazole And 1 M hydrochloric acid to form the buffer. Can you show calculations for the formation of 500 mls of this buffer?

Answer 1

I looked up the pKa for imidazole as 7.00 but you should confirm that in your text or lecture materials. If we call the base BNH2, then
..........BNH2 + HCl ==> BNH3^+ + Cl^-
initial...0.15....0......0..........0
add HCl............x
equil......0.15-x...0....x.........x

Then pH = pKa + log(base)/(acid)
7.30 = 7.00 + log(0.15-x)/x
Solve for x which will be the M HCl you want. I get something like 0.05M
You want 0.500 L of 0.15 buffer so you want 0.075 moles.
0.075 moles imidazole x molar mass = grams.
You can go from here with the amount of HCl to add.
After you finish the calculation take my advice and work the other way to see if adding the materials in your calculation will give you a pH of 7.30

Answer 2

Isn't imidazole c3h4n2 and not bnh2 though?