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Topic: 4 mL of saturated 5.4M NaCl + 2 drops of concentrated 12M HCl  (Read 3448 times)

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Offline gautimagg

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4 mL of saturated 5.4M NaCl + 2 drops of concentrated 12M HCl
« on: November 08, 2015, 10:28:30 AM »
If i put 4 mL of a saturated 5.4 M NaCl solution and then add 2 drops of a concentrated 12M HCl solution to it, how would i balance that? And how do i calculate the change in energy?

What i do know is that when these two mix together the solution becomes cold so this reaction is endothermic. But i do not know where to start to even this all out to get a number going. But im guessing that NaCl seeing as it's an ionic bond its bonds with the HCl and tears it apart making it endothermic.
« Last Edit: November 08, 2015, 10:58:13 AM by gautimagg »

Offline Arkcon

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Re: NaCl + HCl need *delete me*
« Reply #1 on: November 08, 2015, 10:36:24 AM »
Well, you have lots to do, and not all of it is going to be clear.  We're going to have to take this step by step.

If i put 4 mL of 5.4 M NaCl solution and then add 2 drops of 12M HCl solution to it, how would i balance that?

You can start by trying to write a chemical equation from the information that you've been given.  Predicting products is never a simple task, and this reaction, in particular, is going to give us all some problems.  Some context might help, in this case.

Quote
And how do i calculate the change in energy?

Well, a complete and balanced reaction will be needed, but its only the first step.

I hope you don't mind me answering your questions with more questions.  That's what we do here.  Its detailed in the Forum Rules{click}.  We'd like to see your attempt at this problem.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline gautimagg

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Re: NaCl + HCl
« Reply #2 on: November 08, 2015, 10:42:34 AM »
I forgot to mention it as saturated NaCl added with concentrated HCl, i'll edit that in my OP.
I'd think it would look something like this :
NaCl(aq) + conc HCl  :rarrow: NaCl(s) + HCl(aq) 

but what confuses me is the 4mL and 5.4M along with 2 drops and 12M will that affect this in any way or the enthalpy change?


Offline Arkcon

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Re: 4 mL of saturated 5.4M NaCl + 2 drops of concentrated 12M HCl
« Reply #3 on: November 08, 2015, 12:44:27 PM »
That's a better explanation, and a good, clear chemical reaction.  Now you're going to need some formulas handy.  For one, the solubility of NaCl in water, to determine how much NaCl (s) will precipitate.  The production of a solid, from a solution, that will change entropy -- do you know of a formula that relates entropy and enthalpy?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline gautimagg

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Re: 4 mL of saturated 5.4M NaCl + 2 drops of concentrated 12M HCl
« Reply #4 on: November 08, 2015, 04:21:23 PM »
That's a better explanation, and a good, clear chemical reaction.  Now you're going to need some formulas handy.  For one, the solubility of NaCl in water, to determine how much NaCl (s) will precipitate.  The production of a solid, from a solution, that will change entropy -- do you know of a formula that relates entropy and enthalpy?


Are you referring to Gibbs free energy? And if so I've got to admit It's not my strongest suite any good link or direct to a place to learn how to calculate this would be a good help to me!
I'd reckon this but I'm not sure. ∆G= ∆H -T∆S.

Offline Arkcon

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Re: 4 mL of saturated 5.4M NaCl + 2 drops of concentrated 12M HCl
« Reply #5 on: November 10, 2015, 05:24:27 AM »
Not my strongest suit either.  But if you have the Gibbs free energy, you can now relate the enthalpy.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

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