Question

Assuming DIPEA (N,N- diisopropylethylamine) is dissolved in water, write an approximately pH value where:

a. Exactly half of the molecules of DIPEA are protonated (half deprotonated)

b. The ratio of protonated to deprotonated molecules is 100:1 (100 protonated, 1 deprotonated)

c. The ratio of protonated to deprotonated molecules is 1:1000 (1 protonated, 1000 deprotonated)

Thank you!!

Answer 1

Q1.

Apply henderson hasselbach equations

a)

pOH = pKb + log(BH+ / B)

since no pKb data is given, then assume pKb as a value

so...

half molecules protonated, means 1:1 ratio so 1/1 = 1

then

pOH = pKb + log(BH+ / B)

pOH = pKb + log(1)

pOH = pKb

pOH = 14-pH so:

14 - pH = pKb

pH = 14-pKb

b)

for 100:1 ratio then:

pOH = pKb + log(BH+ / B)

pOH = pKb + log(100/1)

pOH = pKb + 2

pOH = 14-pH

14- pH = pKb + 2

pH = 12-pKB

c)

similarly:

for 1:1000 ratio then:

pOH = pKb + log(BH+ / B)

pOH = pKb + log(1/1000)

pOH = 14-pH so:

14-pH = pKb + log(1/1000)

pH = 17 -pKb