Answer:
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 =
