A 1.00-g sample of piperazine hexahydrate is dissolved in enough water to produc

ID: 892132 • Letter: A

Question

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.

pKB1 = 4.22 pKb2= 8.67

A.) What is the initial pH of the solution, before any titrant is added? Answer = 11.23

B.) Calculate the pH at the 25%, 50%, and 75% neutralization points of the first neutralization, respectively. Answer = 10.26,9.78,9.30

C.) What is the pH at the first equivalence point? Answer = 7.56

D.) Calculate the pH at the 25%, 50%, and 75% neutralization points of the second neutralization, respectively. Answer = 5.81,5.33,4.85

E.) What volume of 0.500 M HCl is needed to reach the first equivalence point? Answer = 10.3 mL HCL

F.) What is the pH at the second equivalence point?

I just need someone to show the work for part E and F please!

Molar mass 194.23g/mol

1.00-g sample of piperazine hexahydrate/ 194.23 g/mol = 0.00515 mol

It is a dibase.

E. Reaction is 1:1

0.00515 mol mol piperazine = 0.5 mol/L x V

V= 0.0103 L = 10.3 mL

At the second EP you have the solution of the second conjugated acid having Ka

Ka2= 14 - pKb2= 14 – 8.67 = 5.33

Its concentration is

0.00515 mol / 0.1206 L = 0.0427 mol/L (log 0.0427 = - 1.36)

(the initial volume 100 mL is diluted to 120.6 mL at second EP).

[H+] = (Ka2 c)1/2

pH = (½)pKa2 - (½)logc

= 2.66 + 0.68 = 3.34

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