Question

Calculate the volume (in mL) of a 0.710 M KOH solution that should be added to 4.250 g of HEPES (MW=238.306 g/mol, pKa = 7.56) to give a pH of 7.32.

Answer 1

HEPES is a weak acid lit called HX

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

Hence, the ratio of the two is

[X-]/[HX] = 10^(7.32-7.56) = 0.5754

[X-] = 0.5754[HX]

now write the balanced equation between these two

HX + KOH ---> NaX + H2O

no ofmoles of HX = weight of HEPES / molar mass of HEPES = 4.52 / 238.36 = 0.01896 moles

the equilibrium amount of HX is (0.01896 - x) moles and the equilibrium amount of X- is x; therefore,

x = 0.5754 (0.01896 - x)

1.5754x = 0.010913 moles

x = 0.006927 moles

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

(0.006927 moles) / (0.710 M KOH) = 0.009757L or 9.757 mL