Question

Calculate the volume (in mL) of a 0.620 M KOH solution that should be added to 5.500 g of HEPES (MW=238.306 g/mol, pKa = 7.56) to give a pH of 7.50.

Answer 1

HEPES is a weak acid, so using Henderson-Hasselbach equation:

pH = pKa + log([X-]/[HX])

Hence, the ratio of the two is:

[X-]/[HX] = 10^(pH - pka)

= 10^(7.50 - 7.56)

= 0.251

or

[X-] = 0.251[HX]

Mol of HEPES = mass / molar mass

= 5.5 g / 238 gmol-1

= 0.023 mol

Now consider the following reaction:

HX + KOH ---> KX + H2O

Where, HX = HEPES

the equilibrium amount of HX would be (0.023 - x) moles and the equilibrium amount of X- is x; therefore,

x = 0.251(0.023 - x)
1.251 x = 0.0057 moles
x = 0.0046 moles = mol of KOH

this amount of KOH that would be used to neutralize HEPES, so,

Volume of KOH = (moles) / molarity

= 0. 0046 mol / 0.620 M

= .0074 L or 7.41 mL