Question

You have a 0.025 L solution of piperazine (perhydro-1,4-diazine), a weak dibasic base. The K, values for its conjugate acid are 4.65*10 and 1.86*10-10 (in a solution with an ionic strength of zero). You would like to know the concentration of your solution and decide to titrate with HCl. You make a 0.10 M solution of the acid and use methyl orange as an indicator. You reach the endpoint after adding 54.16 mL of your HCl solution. (a) [The transition range of your indicator is pH = 3.1 - 4.4. Is this a good indicator for your system? Are you seeing an endpoint after your 1st or 2nd equivalence point?] (b) [You ended up using 54.16 mL of your titrant solution, if you were to do this titration again would you use the same concentration of HCI?) (c) How many moles of piperazine were in your initial solution? mol (d) What was the concentration of your initial solution? (e) What was the pH of your initial solution? (f) What was the pH at your first equivalence point? (g) When you have an intermediate species of a polyprotic acid or base (i.e. the monoprotonated form of piperazine) in solution, it can act as a weak acid and as a weak base. Because of this, H' and OH are made by dissociation and hydrolysis in water, respectively, and these two products react with one another. Thus it requires a more systematic treatment to figure out the exact pH of the solution. In general, we can simplify our calculation of pH by using the equation pH m 1(pKantpK2). Redo the calculation for parte with this equation and report the pH you get. pH YpK1+pK2) - (How different are the two pH values? Is this significant?] (h) What was the pH at your second equivalence point?
Dont need (a) or (b) thank you

Answer 1

pKb1 = -Log(4.65*10^-6) = 5.33, pKb2 = -Log(1.86*10^-10) = 9.73

The concentration of salt at first equivalence point = 0.0025 mol/(0.025 + 0.05416) L = 0.0316 M

The concentration of salt at second equivalence point = 0.0025 mol/(0.025 + 2*0.05416) L = 0.0188 M

Part c) Moles of piperazine in the initial solution = 0.025 L * 0.1 mol/L = 0.0025 mol

Part d) Concentration of intitial solution = 0.1 M

Part e) pH of initial solution = 14 - 1/2 (pKb1 - Log0.1) = 14 - 1/2 (5.33 + 1) = 10.83

Part f) pH at first equivalence point = 7 - 1/2 (pKb1 + Log[salt]) = 7 - 1/2 (5.33 + Log0.0316) = 5.08

Part g) pH = 1/2 (14-5.33 + 14-9.73) = 6.47

Part h) pH at the second equivalence point = 7 - 1/2 (pKb2 + Log[salt]) = 7 - 1/2 (9.73 + Log0.0188) = 3