Question

Calculate the volume (in mL) of a 0.790 M KOH solution that should be added to 5.250 g of HEPES (MW=238.306 g/mol, pKa = 7.56) to give a pH of 7.86.

Answer 1

moles of HEPES = 5.25 g / 238.306 g/mol => 0.022 mol

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

Hence, the ratio of the two is:

[X-]/[HX] = 10^(7.86-7.56) = 1.995

or

[X-] = 1.995[HX]

Now consider the following reaction:

HX + KOH ---> NaX + H2O

the equilibrium amount of HX is (0.022 - x) moles and the equilibrium amount of X^- is x; therefore,

x = 1.995(0.022 - x)

x = 0.04389 - 1.995x
2.995x = 0.04389 moles
x = 0.01465 moles

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

[(0.01465 moles ) / (0.790 moles KOH) ]*1000 mL= 18.55 mL