# Question bfeb7

Apr 3, 2015

I'm a little confused by the data given, and here's why.

Ethylenediamine (en), or ${\left(C {H}_{2}\right)}_{2} {\left(N {H}_{2}\right)}_{2}$, is a weak base, which implies that you've essentially performed a weak base - strong acid titration.

Because en is dibasic, you'll have 2 equivalence points on the titration curve, and the pH after neutralization 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 $\text{HCl}$ solution used is far too little.

A $\text{30% v/v}$ 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 density 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

$C = \frac{n}{V} \implies n = C \cdot V$

n_("HCl") = "5 M" * 2.5 * 10^(-3)"L" = "0.0125 moles HCl"

In this case, It'sclear that you've used too little $\text{HCl}$ and neutralization is a long way ahead, i.e. you need more $\text{HCl}$.

Another approach I'd take is to work backwards from the number of moles of $\text{HCl}$ to get the number of moles of en.

The second equivalence point requires 2 times more moles of $\text{HCl}$ 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.