Piperazine, HN(C4H8)NH, is a diprotic weak base used as a corrosion inhibitor an
ID: 877536 • 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 1.00-g sample of piperazine hexahydrate is dissolved in enough water to produce 100.0 mL of solution and is titrated with 0.500 M HCl.
Part A
What is the initial pH of the solution, before any titrant is added?
Express your answer numerically with two digits after the decimal point.
Part B
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.
Part C
What is the pH at the first equivalence point?
Part D
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.
Explanation / Answer
C4H10N26H2O molar mass = 194.23 g / mol
molarity = ( weight / molar mass ) x 1000 / V
= ( 1 / 194.23 ) x 1000 / 100
= 0.0515 M
Part A ) : initial pH
pOH = 1/2 (pKb1- logC)
= 1/2 (4.22 -log 0.0515)
= 2.75
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 2.75
pH = 11.25
Part B )
(i) 25 % neutralisation point
Pip + HCl --------------------> PipH+ + Cl-
0.0515 0.5 0 0 -------------initial
0.0515-x 0.5-x x x --------------equilibrium
0.0387 0.487 0.0128 0.0128 --------------equilibrium
calculation of x :
x= 25% of 0.0515
= 25 x 0.0515 / 100
= 0.0128
here base , salt remained in the solution. it is basic buffer
pOH = pKb1 + log [salt / base]
= 4.22 + log (0.0128/0.0387)
= 3.74
pH + pOH = 14
pH = 10.26
(ii) 50 % neutralisation point
Pip + HCl --------------------> PipH+ + Cl-
0.0515 0.5 0 0 -------------initial
0.0515-x 0.5-x x x --------------equilibrium
0.0257 0.473 0.0257 0.0257
calculation of x :
x= 50% of 0.0515
= 50 x 0.0515 / 100
= 0.0257
pOH = pKb1 + log [salt/base]
pOH = 4.22 + log (0.0275/0.0257)
pOH = 4.22
pH = 9.78
(iii) 75 % neutralisation point
Pip + HCl --------------------> PipH+ + Cl-
0.0515 0.5 0 0 -------------initial
0.0515-x 0.5-x x x --------------equilibrium
0.0129 0.461 0.0386 0.0386
calculation of x :
x= 75% of 0.0515
= 50 x 0.0515 / 100
= 0.0386
pOH = pKb1 + log [salt/base]
pOH = 4.22 + log (0.0386/0.0129)
pOH = 4.69
pH = 9.30
remaining problem solve similar way above shown. iam sorry it is very lengthy problem . i could solved these only.