Calculate the volume in mL of a 0.250 M KOH solution that should be added to 5.250g of HEPES (MW=238.306g/mol and pKa=7.56) to give pH of 7.88

Question

Calculate the volume in mL of a 0.250 M KOH solution that should be added to 5.250g of HEPES (MW=238.306g/mol and pKa=7.56) to give pH of 7.88.

KOH volume = mL

Answer 1

HEPES is a sulfonic acid derivative. We can represent it as a weak acid, HA. When irate with KOH, we form a buffer containing the weak acid HA and the conjugate base , A-

HA + KOH ==> H2O + A-

Henderson Hasselbalch eq:

pH = pKa + log [A-] / [HA]

7.88 = 7.56 + log [A-] / [HA]

Log [A-] / [HA] = 0.32

[A-] / [HA] = 2.089

Initial moles HEPES = 5.250 g x 1 mol/238.3 g = 0.02203 moles

x / 0.02203 - x = 2.089

x = 0.01489 moles A-

moles HA remaining = 0.02203 - 0.01489 = 0.00714 moles HA

To produce 0.01489 mols A- how many mls of 0.350. M KOH do we need?

0.01489 mol x 1L/0.250 mol = 0.05956 L = 59.56 mls

Be sure to check all the math.