Does the volume of water change after a neutralization reaction ?

[B4ssHunter]

suppose you have 2 litres of pure hcl in a solution of 4 litres of radioactive h2o , so the concentration of HCl is 1/3 of the whole solution , now suppose you neutralized that solution with Sodium hydroxide , what happens to the radioactive solvent ? is it involved in the reaction ? does its concentration change ?

 

[SteamKing]

Write out the chemical equation of the neutralization reaction. Analyze it. Do the water molecules from the solvent react chemically with the acid or base?

 

[Borek]

To clarify: do you mean volume of water, or volume of the solution?

What do you mean by "2 litres of pure HCl"? Gaseous? Liquid at -85°C?

Besides, I wonder what you mean when you write "radioactive H₂O". The only way to make water radioactive is to either replace hydrogen or oxygen with one of their radioactive isotopes, or to dissolve something radioactive - then we have no "radioactive water", but "radioactive solution".

This may be all a nitpicking, but from your question it looks like you are missing the differences between all those things.

 

[B4ssHunter]

my apologies , i meant liquid HCl , and by radioactive H2O i meant H2O with radioactive Oxygen isotope , anyway this is not the problem .
my question is
if i have a solution of HCl , for the sake of simplicity let's say it is dissolved in some other solvent , anything other than water , and HCl reacts with NaOH , what will happen to that solvent ?
i know nothing should happen to it and it should come clean out of the reaction , but my chemistry book does not agree , i do not want to post the example here because i don't want to be a homework question , but is it possible that the solvent gets consumed in this reaction ? * let the solvent be ethanol *

 

[Yanick]

Your question is a bit confusing. Can you post the question and/or the relevant text?

 

[B4ssHunter]

basically the question is
i have a solution** of HCl * say 3 litres * it reacted with 1 mole of calcium carbonate * so 2 moles of HCl reacted with CaCO3 * , some of the HCl remained , the volume of the solution after the reaction with CaCO3 is 1 Litre , measure the concentration of HCl .
so basically what i am supposed to do is to find the volume of HCl that reacted with the carbonate by subtracting , 3-1 = 2 litres , then divide the number of moles by volume to get the concentration .
but it just doesn't make sense , the volume consumed is the volume of pure HCl , so the concentration i deduce should be the concentration of Pure HCl , not the concentration of HCl in the solution , if not , then if enough CaCO3 was supplied , the whole solution would be depleted , and only what remains should be the H20 and Calcium Chloride released from the reaction . and the solvent would magically disappear i suppose * unless i am wrong *

 

[Borek]

I don't think you mean liquid HCl (nor pure HCl), I suppose you mean a solution of HCl of a known concentration (so the water is already there). At room temperature HCl is a gas, it becomes liquid at -85°C (assuming 1 atm pressure).

Alternatively - yes, you mean pure HCl, but that's because you have no idea it doesn't make sense, and/or you are misreading the book which speaks about HCl solution all the time.

You are mixing moles, litres, volumes, everything together as if these things were equivalent. They are not.

 

[B4ssHunter]

there must be somekind of a misunderstanding , first of all it doesn't matter if the HCl was liquid or gas , its dissolved in a solution , the way i was given the second volume was not direct , anyway let's not speak about moles for a while , the volume of the solution at first was 3 litres , after the reaction , the volume became 1 litre * note that some HCl remained * , wouldn't that mean that the volume of The Pure HCl that was disolved in the solution = 2 litres ? please correct me if i am wrong
--
oops i made a mistake in the post above , i did not mean a volume of HCl , i meant a solution of HCl , i am truly sorry for that stupid mistake

 

[Borek]

2 L of solution of HCl and 2 L of pure HCl are completely different things.

Now that we have established we are talking about a solution, what is the concentration of the dissolved HCl?

 

[B4ssHunter]

thats what i am supposed to find
i will post the exact numbers in the problem
the volume of the solution is 50 mL , the solution reacted with 0.0101 moles of CaCO3 , * so 0.0202 moles of HCl reacted with the carbonate * , the remaining solution needs 24.6 mL of a specific NaOH solution
25 mL of that NaOH solution would neutralize 23.7 mL of our HCl solution
so basically we deduce the volume of the HCl using cross multiplication
25 mL neutralize 23.7 mL
then 24.6 mL would neutralize X mL
which happened to be 23.32 mL
so the volume remaining i suppose is 23.32 mL , now to find the concentration i subtract this volume from the initial volume to get the volume of HCl that reacted with the carbonate , then divide the number of moles ( 0.0202 / 23.32 * 10-3) to get the concentration ,.
what doesn't make sense to me , is that the volume subtracted from the solution is the volume of pure HCl , not HCl in the solution
if i could figure out the concentration that way, then this means using enough CaCO3 i could totally diminish the whole solution and all what remains is the H20 and salts released from the neutralization reaction * this is why i said radioactive h20 from the beginning , to distinguish between the solvent and water released from the neutralization reaction *

even if we had used another initial HCl solution with different concentration , we would still get a decrease of 23.23 mL when 0.0202 moles of HCl reacts with CaCO3 * under constant Temp and pressure *

 

[Borek]

Subtracting volumes is a dangerous shortcut, especially when you don't understand what you are doing. You don't have "volume of HCl", you have "volume of HCl solution". This volume contains some number of moles of HCl. After the reaction it contains smaller number of moles of HCl, not smaller volume of HCl.

You have still failed to explain what is the problem you are solving:

the volume of the solution is 50 mL , the solution reacted with 0.0101 moles of CaCO3 , * so 0.0202 moles of HCl reacted with the carbonate * , the remaining solution needs 24.6 mL of a specific NaOH solution
25 mL of that NaOH solution would neutralize 23.7 mL of our HCl solution

I don't see a question here, just the data.

Is it that hard to quote the problem entirely, to make things clear? You are giving us a piece of information at a time, wasting our and your time for completely unnecessary guesses.

I am moving the thread to homework section.

 

[B4ssHunter]

i am asked to find the concentration as i have stated in the beginning of my post
my question is , is it possible to deduce the concentration from the volume of the solution that was consumed ? the volume of the solution did decrease * due to the reaction with CaCO3 * , i don't think this should happen but that's how the problem is , can i just deduce the concentration from the volume lost during the reaction with caco3 and the number of moles that reacted with CaCO3

 

[Borek]

i am asked to find the concentration as i have stated in the beginning of my post

Concentration of what?

If you will not quote whole problem word for word I am going to leave the thread. You are wasting my time making me guess what the problem is. It is up to you to present the problem in a clear, unambiguous way.

the volume of the solution did decrease

I don't think so. As I said before - you are misunderstanding and/or misinterpreting what the question says.