Question

Calculate the volume (in mL) of a 0.700 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 7.76.

Answer 1

Since HEPES (HX) is a weak acid, you can use the Henderson-Hasselbach equation:

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

7.76 = 7.56 + log([X^-]/[HX])

log([X^-]/[HX]) = 0.20

([X^-]/[HX]) = 10^0.20

([X^-]/[HX]) = 1.58

[X-] = 1.58 [HX]

Molar mass of HEPES = 238.306 g/mol

So, 238.306 g of HEPES = 1 mol

4.5 g of HEPES = (4.5 / 238.306) mol = 0.019 mol

Now consider the following reaction:

HX + KOH NaX + H2O

the equilibrium amount of HX is (0.019 - x) moles and the equilibrium amount of X^- is x;

Since,

[X^-] = 1.58 [HX]

x = 1.58 (0.019 - x)

x = 0.03 – 1.58x)

2.58x = 0.03

x = 0.03/2.58 = 0.012 moles

This amount is exactly the amount of KOH that was used to neutralize HEPES. Thus the volume of KOH(aq) was:

(0.012 moles) / (0.700 M KOH) = 0.0084 L or 8.4 mL