Question

Calculate the volume (in mL) of a 0.360 M KOH solution that should be added to 4.500 g of HEPES (MW=238.306 g/mol, pKa = 7.56) to give a pH of 7.50.

Answer 1

Using Henderson-Hasslebalch equation-

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

7.50 = 7.56 + log[A^-]/[HA]

[A^-]/[HA] = 10^7.50-7.56 = 0.871

[A^-]/[HA] = 0.871

[A^-] = 0.871[HA]

moles of HEPES = 4.5g/238.306g/mol = 0.0189mol

HA + KOH ---> A^- + H2O

so, let equilibrium concentration of A^- = x

equilibrium conc of HA = 0.0189 -x

so,

x = 0.871(0.0189-x)

1.871x = 0.01646

x= 0.00880moles

so, A^- = 0.00880moles

this is the amount of KOH needed to neutralize HEPES-

so,

Volume of KOH = moles/Molarity = 0.00880moles/0.360mol/L = 0.0244L or 24.44mL