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QUESTION

# How do you calculate the weight of ethylenediamine (en) used? (note: 30% en solution (15mL) was neutralised with 5M HCl (2.5mL), density of en also given to aid in calculation :0.90g/mL)

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.