Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor an
ID: 1013617 • Letter: P
Question
Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor and an insecticide and has the following properties:
pKb1=4.22pKb2=8.67
For writing the reactions of this base in water, it can be helpful to abbreviate the formula as Pip:
Pip+H2OPipH++H2OPipH++OHPipH22++OH
The piperazine used commercially is a hexahydrate with the formula C4H10N26H2O.
A) The percent neutralization points for a titration indicate the percent of the reagent that has reacted. For example, at the 25% neutralization point for piperazine, 75% (0.75×[Pip]) of the base remains, and 25% is neutralized to form PipH+.
Calculate the pH at the 25%, 50%, and 75% neutralization points of the first neutralization, respectively.
Enter the pH at the 25%, 50%, and 75% neutralization points to two decimal places separated by commas.
B) What is the pH at the first equivalence point?
Express your answer numerically with two digits after the decimal point.
C) The percent neutralization points for a titration indicate the percent of the reagent that has reacted. For example, at the 25% neutralization point for PipH+, 75% (0.75×[PipH+]) of the base remains, and 25% is neutralized to form PipH22+.
Calculate the pH at the 25%, 50%, and 75% neutralization points of the second neutralization, respectively.
Enter the pH at the 25%, 50%, and 75% neutralization points to two decimal places separated by commas.
Explanation / Answer
A) Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
at 25% neutralization,
pH = (14 - 8.67) + log(75/25) = 5.81
at 50% neutralization,
pH = (14 - 8.67) + log(50/50) = 5.33
at 75% neutralization,
pH = (14 - 8.67) + log(25/75) = 4.85
pH = 5.81,5.33,4.85
B) pH at equivalence point
pH = 1/2(pKa1+pKa2) = 1/2(5.33 + 978) = 7.55
C) For second neutralization point,
at 25% neutralization,
pH = (14 - 4.22) + log(75/25) = 10.26
at 50% neutralization,
pH = (14 - 4.22) + log(50/50) = 9.78
at 75% neutralization,
pH = (14 - 4.22) + log(25/75) = 9.30
pH = 10.26,9.78,9.30