Question

PIPES buffer has a pKa of 6.76 at 25.0oC and 6.6 at 37.0oC.

What is the enthalpy of the acid-base reaction?

What is the pKa of this buffer at 4.0oC?

Show steps and I will give good rate!

Answer 1

T1 = 25 + 273 = 298 K

T2 = 37 + 273 = 310 K

pKa1 = 6.76

Ka1 = 10^-6.76 = 1.74 x 10^-7

pKa2 = 6.6

Ka2 = 10^-6.6 = 2.51 x 10^-7

ln (Ka2 / Ka1 ) = ( H / R ) (1/T1 -1/T2)

ln (2.51 x 10^-7 /1.74 x 10^-7 ) =(H / 8.314 x 10^-3 ) [1/298 -1/310]

H = 92.40 kJ / mol

Enthalpy = H = 92.40 kJ / mol --------------------> answer

T3= 273 + 4 = 277 K

ln (Ka1 / Ka3 ) = ( H / R ) (1/T1 -1/T2)

ln (1.74 x 10^-7 / Ka3 ) =(92.40 / 8.314 x 10^-3 ) [1/277 -1/298]

Ka3 = 1.03 x 10^-8

pKa = -log Ka3

= -log (1.03 x 10^-8 )

= 7.99

pKa = 7.99 ------------------------> answer