Question

An experiment requires a HEPES buffer,

Answer 1

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 $\frac{HEPES}{HEPES-H^+}$

2,818 =  $\frac{HEPES}{HEPES-H^+}$ (1)

Knowing you need to make 1,0L×0,10M = 0,1 moles of HEPES

0,1 = [HEPES] + [HEPES-H⁺] (2)

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×$\frac{260,3g}{1mole}$ = 26,03 g of HEPES

0,02619 moles of HCl÷6,0M HCl = 4,365x10⁻³ L ≡ 4,365 mL HCl 6,0M

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!