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!

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.