Answer:
To make 1,0L of 0,10 M buffer of HEPES pH = 8,0 you need to add 26,03 g of HEPES and 4,365 mL of HCl and water until 1,0L
Explanation:
The dissociation of HEPES buffer is:
HEPES-H⁺ ⇄ HEPES + H⁺
The effective buffer range is pk±1. Thus: The effective buffer range is (6,55-8,55)
To prepare a buffer of pH = 8,0 you need to use Henderson-Hasselbalch formula, thus:
8,0 = 7,55 + log₁₀
2,818 = <em>(1)</em>
Knowing you need to make 1,0L×0,10M = <em>0,1 moles of HEPES</em>
0,1 = [HEPES] + [HEPES-H⁺] <em>(2)</em>
Replacing (2) in (1):
HEPES-H⁺ = 0,02619 moles
Thus:
HEPES = 0,07381 moles
The other equation you need to use is:
HEPES + HCl → HEPES-H⁺ + Cl⁻
Thus, to obtain the buffer of pH = 8,0 you need to add 0,1 moles of HEPES and 0,02619 moles of HCl:
0,1 moles HEPES× = <em>26,03 g of HEPES</em>
0,02619 moles of HCl÷6,0M HCl = 4,365x10⁻³ L ≡ <em>4,365 mL HCl 6,0M</em>
Thus, to make 1,0L of 0,10 M buffer of HEPES pH = 8,0 you need to add 26,03 g of HEPES and 4,365 mL of HCl and water until 1,0L
I hope it helps!