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 7.60.

Answer 1

Since HEPES is a weak acid, We should use the Henderson-Hasselbach equation:

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

Hence, the ratio of the two is:

[X-]/[HX] = 10^(7.60-7.56) = 1.09

or

[X-] = 1.09[HX]

Now consider the following reaction:

HX + KOH ---> NaX + H2O

mole of HEPES
4.5/238.306 = 0.0188 mol
the equilibrium amount of HX is (0.0188 - x) moles and the equilibrium amount of X- is x; therefore,

x = 1.09(0.0188 - x)
2.09x = 0.02049 moles
x = 0.0098 moles

this amount is exactly the amount of KOH that was used to neutralize HEPES, thus the volume of KOH(aq) was:

(0.0098 moles) / (0.360 M KOH) = 0.0272 L or 27.0 mL