I’m a little confused by the data given, and here’s why.
Ethylenediamine (en), or ##(CH_2)_2(NH_2)_2##, is a weak , which implies that you’ve essentially performed a weak base – strong acid .
Because en is dibasic, you’ll have 2 equivalence points on the titration curve, and the after will be less than 7.
Now, if you’ve used 15 mL of a 30% en solution in the titration, the 2.5 mL volume of ##”HCl”## solution used is far too little.
A ##”30% v/v”## solution would have 30 mL of en in every 100 mL of solution; if you’ve used 15 mL of solution, then the volume of en is
##”30%” = V_(“en”)/V_(“solution”) * 100 => V_(“en”) = (30 * V_(“solution”))/100##
##V_(“en”) = (30 * “15 mL”)/100 = “4.5 mL”##
Using its given will give you the mass
##4.5cancel(“mL”) * “0.90 g”/(1cancel(“mL”)) = “4.05 g”##
Here’s what doesn’t seem right to me. In this case, the number of moles of en would be
##4.05cancel(“g”) * “1 mole”/(60.10cancel(“g”)) = “0.0674 moles en”##
The first equivalence point would have the number of moles of en equal to the number of moles of HCl, which is
##C = n/V => n = C * V##
##n_(“HCl”) = “5 M” * 2.5 * 10^(-3)”L” = “0.0125 moles HCl”##
In this case, It’sclear that you’ve used too little ##”HCl”## and is a long way ahead, i.e. you need more ##”HCl”##.
Another approach I’d take is to work backwards from the number of moles of ##”HCl”## to get the number of moles of en.
The second equivalence point requires 2 times more moles of ##”HCl”## than of en, which means that
##n_(“en”) = “0.0125 moles”/2 = “0.00625 moles en”##
This is equivalent to
##0.00625cancel(“moles en”) * “60.10 g”/(1 cancel(“mole en”)) = “0.376 g en “##, or
##rho = m/V => V_(“en”) = m/(rho) = (0.376cancel(“g”))/(0.90cancel(“g”)/”mL”) = “0.42 mL”##
which corresponds to
##V_(“solution”) = (V_(“en”) * 100)/30 = “1.4 mL en solution”##
As I can see it, you’ve either used 1.5 mL of solution instead of 15 mL, or 25 mL of HCl instead of 2.5 mL.
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