Calculate the volume, in milliliters, of a 0.470 M KOH solution that should be added to 5.250 g of HEPES (MW = 238.306 g/mol, pKa = 7.56) to give a p

17.13mL

Explanation:

Complete question:

to give a pH of 7.32.

Hepes is a weak acid. When HEPES reacts with KOH its conjugate base, HEPES⁻ is produced:

HEPES + KOH → HEPES⁻

Now when you have in solution both the weak acid and conjugate base a buffer is produced. The equation that relates pKa and pH of a buffer is H-H equation:

pH = pKa + log [HEPES⁻] / [HEPES]

Where [] could be taken as the moles of each specie.

Replacing:

7.32 = 7.56 + log [HEPES⁻] / [HEPES]

-0.24 = log [HEPES⁻] / [HEPES]

0.57544 =  [HEPES⁻] / [HEPES] (1)

Now, moles of 5.250g of HEPES is:

Mass HEPES:

5.250g * (1mol / 238.306g) = 0.02203 moles HEPES.

That means:

0.02203 moles = [HEPES⁻] + [HEPES] (2)

Replacing (2) in (1):

0.57544 =  0.02203 moles - [HEPES] / [HEPES]

0.57544[HEPES] =  0.02203 moles - [HEPES]

1.57544[HEPES] = 0.02203 moles

[HEPES] = 0.01398 moles HEPES.

And [HEPES⁻] = 0.02203 moles - 0.01398 moles =

0.00805 moles of HEPES⁻

As HEPES⁻ is produced from HEPES and KOH, if we add 0.0805 moles of KOH, we will have in solution 0.00805 moles of HEPES⁻ and 0.01398 moles of HEPES.

To obtain 0.00805 moles of KOH from a 0.470M KOH, you need to add:

0.00805 moles KOH * (1L / 0.470mol) = 0.01713L of 0.470M KOH =

it is reliable if two scientists recorded it