A-level Organic Chemistry

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I would say that you ought to be off LI and carrying on with your coursework, but I'm in need of help yet again.

- A sample of ammonia was dissolved in water to produce 100 cm3 of a solution, X. 10 cm3 of this solution was made up to a volume of 250 cm3. 25.0 cm3 of this diluted ammonia solution was then titrated with aqueous hydrochloric acid of concentration 0.110 M. 37.10 cm3 of the acid was required to neutralise the ammonia solution.

Write and equation for the reaction between ammonia and HCl, and calculate the number of moles of ammonia in solution X. Calculate the concentration of solution X in moldm^-3.

As for the equation, this is what I said:

NH3 + HCl = NH4+ + Cl-

Then I said that once mole of acid reacts with 1 mole of ammonia.
Thus the calculated moles of acid = the number of moles of acid. (for 25cm3 of the ammonia solution)
Then for 250 cm3, I multiplied it by 10. Now I'm lost as to how to continue.
 
Wow, that's a scary-looking question...okayy, you're right in saying the moles of acid = moles of ammonia. You can find the concentration of the 25cm3 of ammonia by doing conc = mols/volume. From what I understand, this concentration is equal to the concentration of the 250cm3 (you're just taking a 25cm3 sample of that 250cm3 of solution for the titration.
However, this 250cm3 is obtained by diluting 10cm3 of X, i.e. diluting X by 25 times. The concentration of the original solution is therefore 25 times the concentration calculated for the 250cm3. Hope that makes sense...

(Btw, I've hardly ever used the unit, so I was wondering, is 0.110 M simply equivalent to 0.110 mol/dm3??)
 
^Hmmm. I may need some elaborate explanation... But I'll ask for that later.

And as far as my knowledge goes, yes, M = mol/dm3.
 
I'd go about it like this:
you have X moles in 100mL, in 10mL there's 0.1X moles (because you equally divided th solution so 1 tenth of X to every 10mL), in 250mL still 0.1mL(because you just added water), then you divided it by 10 again, so in 25mL tehre's 0.01X moles.

0.01X =n(Hcl) = c(HCl)×V(HCl)
X = 100×0.110 M×0.0371L = 0.4081 moles

ml = cm3
L = dm3
 
^ I completely agree. So if there's 0.4081 mols in 100cm3, there's 4.081 mols in 1 dm3 (i.e. the concentration is 4.081mols/dm3). I think I obtained the same.
 
Ah... I just checked my working out the first time I did it. And it seems I got the same answer. Just wasn't sure if it was right.
 
Is there any reason why ethanolic solutions are used for particular reactions?
 
The Ruler said:
Is there any reason why ethanolic solutions are used for particular reactions?
Click to expand...

I'm not sure but they're used in the lipid emulsion test arn't they? To break up tryglycerides (esters) into fatty acids (carboxylic acids) and gycerol (alcohol).

So basically for hydrolysis for this particular reaction.
 
^Hmmm... Yes, I know. But I don't see any water when a halogenoalkane reacts with KCN or ammonia.

- This was in reply to wtp's post.
 
I think basically, if there's any water present, different products are formed (the reaction of the halogenoalkane with the water itself becomes one of the steps of the overall reaction). Depending on what you wanna get, you use an ethanolic solution instead of an aqueous solution (the ethanol won't react with anything as easily...)
 
You need a medium/solvent in which the reaction (between the halogenoalkane and ammonia, for example) can take place. The solvent that'd usually be used is water, but its not suitable in this case, because the water itself would react with the halogenoalkane. So, the alternative is ethanol...
 
Sodium Oxide is a basic oxide.
Write the equation for the reaction of sodium oxide with water

I said: Na2O + H2O ----> 2NaOH

Then, state the type of bonding in sodium oxide and explain how particles react to form the product in the above equation.

Type of bonding: Ionic. But I don't exactly understand what the second part of the question is looking for.

EDIT: Is this equation correct - Al2O3 + 2NaOH ---> 2NaAlO2 + H2O
 
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u shud get JCR A level Briggs, helped me a lot for my A level.
i dont think its corect, both are base, it shud be a displacement, well i think so!!! u shud check it
 
Hmmm... I haven't really done this topic, but this might be helpful: Chemguide - Period 3 Oxides. It gives the following equation for the reaction of aluminium oxide with (aqueous) sodium hydroxide...
Al2O3 + 2NaOH + 3H2O --> 2NaAl(OH)4
(Btw, aluminium oxide is what is called an amphoteric compound...it reacts as both an acid and as a base)

As for how the first reaction occurs, I'd probably start by saying how the sodium oxide dissociates in solution into the sodium & oxide ions (and of course, water dissociates into H+ and OH- ions).

According to chemguide, lol, oxide ions are strong bases and have a "high tendency to combine with hydrogen ions", and thereby form another hydroxide ion.

At the same time, the sodium ions will combine either with the hydroxide ions dissociated from water, or one of the hydroxide ion formed in the step above, to form the final product, sodium hydroxide...atleast, that's how I imagine it :><:...hope that helps!

Edit: I was just thinking, and wanted to add an equation (I use the term loosely; I doubt it's ever written like this) which may or may not make things clearer...

Na2O + H2O --> 2(Na+) + (O2-) + (H+) + (OH-).......[Ions dissociate]
(O2-) + (H+) --> (OH-)
2(Na+) + (OH-) + (OH-) --> 2(Na+) + 2(OH-) --> 2NaOH

Lol, superscripts would be nice...
 
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