Question

calculate the volume (in mL) of a .310 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 7.28

Answer 1

HEPES is a weak acid and KOH is a strong base

therefore, expect a buffer formation

the henderson hasselbalch equation:

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

then, the ratio of the two is:

[X-]/[HX] = 10^(7.54-7.28) = 1.8197

so

[X-] = 1.8197 [HX]

Now consider the following reaction:

HX + KOH ---> NaX + H2O

for each addition of base, there is creation of conjugate and reaction of acid

the equilibrium amount of HX is (0.02517 - x) moles and the equilibrium amount of X- is x; therefore

x = 1.8197(0.02517 - x)

x = 0.045801 - 1.8197x

2.8197x = 0.045801

x = 0.045801 /2.8197

x = 0.016243 moles

therefore:

M = mol/V

V = mol/M

so

(0.016243moles) / (0.31 M KOH) = 0.05239 L or 5.23 mL