An experiment requires a hepes buffer,
ph 8.0. write an equation for dissociation of hepes.
what is its effective buffer range? describe how you
would make 1.0 l of 0.10 m hepes buffer (mw = 260.3
g/mol) using 6.0 m hcl. pk = 7.55

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 =  (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× = 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!

2b2t

explanation:

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