Calculate the pH at the 25%, 50%, and 75% neutralization points of the second ne

Question

Calculate the pH at the 25%, 50%, and 75% neutralization points of the second neutralization, respectively

Info:

Piperazine,HN(C4H8)NH , a diprotic weak base has the following properties: pKb1 = 4.22, pKb2= 8.67

For writing the reactions of this base in water,abbreviate the formula as Pip:

Pip+H2O-->/<--PipH+OH-, PipH+ + H2O-->/<-- PipH2+ + OH-

The piperazine used commercially is a hexahydrate with the formula C4H10N2 * 6H20.

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 .

Explanation / Answer

This is exactly the same as what we did in (b), except the titration reaction is now PipH+ + H+ ==> PipH2 2+. Be sure to use pKa2 = 14.00 - pKb2 = 14.00 - 8.67 = 5.33. The moles of PipH+ and PipH2 2+ are exactly as they were in (b). You should get these results: pH at 25% = 5.81 pH at 50% = pKa2 = 5.33 pH at 75% = 4.85

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