 2:22 PM
4  For a lab assignment, I combined 10mL of 1M sodium sulfate solution with 10mL of 1M calcium chloride solution. The ionic equation I came up with is: \$\$\ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + 2Cl^- (aq) -> CaSO4(s) + 2Na^+ (aq) + 2Cl^- (aq)}\$\$ I was then asked in a lab question the following... Isn't your balanced molecular equation supposed to be : \$\ce{Na2SO4(aq) + CaCl2(aq) -> CaSO4(s) + 2NaCl(aq)}\$ ? @LievenB That's what I have for when I use a 1M solution of calcium chloride, but if I use 1M chloride ion solution and 1M calcium ion solution, I think I'm supposed to have that molecular equation instead. Could it be that you are confusing concentrations and reaction equations? : only a small amount of the two solutions will react (when using 1M chloride ion solution and 1M calcium ion solution) and the other part won't react : you can't have free sodium ions without any counterion (such as Cl) in a solution...? @LievenB You can't? what do you mean? 2:22 PM
@LievenB Can you not have free (sodium) ions in a solution? My question is with free calcium ions and chloride ions, not free sodium ions (the other reactant, sodium sulfate, is not changed as far as I can tell the question implies). No, not free sodium ions without a counterion - imagine evaporating all the water in the solution, then what would the sodium ion do? Wouldn't they just be left over? No, the sodium ion will try to find its counterion and then reconnect to it..
such as Na+(aq) + Cl-(aq) -> evaporate water -> NaCl
I think that the available CaCl2 will react with the available Na2SO4 to then completely react to CaSO4 and Na2SO4
but the extra CaCl2 molecules (which provide the extra Cl ions) will not further react Is it literally impossible to place only sodium ions in water, then evaporate the water? Yes - I asked some kind of similar question a few months ago : chemistry.stackexchange.com/questions/6932/… 2:32 PM
Okay, then what did my lab question mean by "If the reaction called for 1M chloride ion with 1 M calcium ion?" Lets look at 1M chloride ion :
this means that a certain amount of CaCl2 is dissolved, yielding 1M Cl- ions and also other Calcium ions
because free ions, as mentioned earlier can't exist without their counterion
(and here so : chlorine ions can't exist without the calcium counterions)
(this is 1M chloride ion added in the beginning)
And I think that "called for" means produced ions
then I guess that you will have a final solution containing 1M Cl- ions and 1M Ca2+ ions (with their counterions: sodium and sulfate, respectively)
______________________ 2:48 PM
I'm not really following. Does that mean my reactant is CaCl^+ solution? yes So, I still have a solution with a net charge. What makes that different?
And besides changing my reactants from Ca(aq) and Cl(aq) to CaCl(aq), what does that change in my equation? First : you need to use CaCl2(aq)
When you have a solution containing CaCl2,Na2SO4
then they will always
react following this reaction equation:
Na2SO4(aq)+CaCl2(aq)->CaSO4(s)+2NaCl(aq) (molecular equation)
So by adding a larger/smaller concentration
of CaCl2 or NaCl
then the reaction equation will still be the same
it is just the amounts, the concentrations, that will be different
So if you have 1mole of Na2SO4
and 1 mole of CaCl2 That's the equation I had initially for 1M Na2SO4 and 1M CaCl2. then all reactants will react
but when you add a larger concentration (eg 2moles of CaCl2 and only 1 mole of Na2SO4)
then you will still have 1 mole of CaCl2
that will not have reacted
So your first equation is correct
(2Na+(aq)+SO2−4(aq)+Ca2+(aq)+2Cl−(aq)->CaSO4(s)+2Na+(aq)+2Cl−(aq)) 2:57 PM
Right, I know that But the equation does not change following differences in the concentrations Sure.
So then, what do they really mean by "If the reaction called for 1M chloride ion with 1 M calcium ion"? What are the substances there?
My initial equation is 2M chloride ion with 1M calcium ion, I think... yes
but does 'called for' mean the amounts after the reaction or before
?
Because if you add 1M chloride ions and 1M calcium ions to the solution in the beginning
then you will have a complete reaction I'm not 100% certain, but I believe they meant in the reactants, i.e. before yeah I'm not that certain too, but I agree with you 3:02 PM
i.e. Ca^2+ and Cl^- not Ca^2+ and 2Cl^- no, the Ca2+ and 2Cl- are correct, we don't need to change the reaction euation
equation
but we know that,
by looking at the reaction equation
that
tells us that 1M of CaCl2
reacts with 1M of Na2SO4
So in case we have 1M of both reactants, then the reaction
will be complete Yeah, but that's 1M of calcium ion and 2M of chloride ion
Or am I wrong? yes
(i mean you are right with that)
So then that means adding 1M of CaCl2 Yep. And if I start with 1M of calcium ion and 1M of chloride ion, I can't get a full mole of CaCl2. Only 0.5 moles of it; then 0.5moles of chloride ion left over. That's my understanding i see
so there is no mention what your source of Cl- ions is(the compound?)
or wait
Can we also see the question as follows:
If the reaction called for 1M chloride ion with 1 M calcium ion, would you still use 10 mL of each solution for a complete reaction? Explain your reasoning.
where the 'reaction called for' means that the reaction itself needs 1M cl-ions and 1M Calcium ions to achieve a complete reaction? 3:16 PM
Do you mean omit the sodium sulfate entirely? I'm not sure This is why I asked whether it was possible to have a solution without a counterion - if I could put in 20mL of 1M chloride ion solution, then no worries. yeah - I see...
If they mean the same reaction equation, then I think that your answer would be no - because the reaction needs 1M CaCl2 to achieve complete reaction with Na2SO4, and so needs 2M of Cl-ions
but in this case there aren't 2M of Cl-ions so I would suspect that there will be no complete reaction So, going back to my overarching question: "what is a complete reaction?"
Is a complete reaction one where all reactant ions formed new compounds? Yes, when the solution has fully reacted 3:22 PM
I linked to a few definitions in my question - is the one from answers.com correct, and the others incorrect? wait - I think that the other ones are correct (I would trust the University&..)
I'm just looking in to this...
So : "A complete reaction is one where all of at least one of the available reactants is used up and converted into products."
So in order to use up one of the two compounds (Na2SO4 or CaCl2), and we know that the reaction equation tells us that if 1M of both molecules are present, then both compounds have reacted. So 2M chloride ions (=1M CaCl2) added in the beginning solution (just like you did in the lab) will achieve a complete reaction...
If we would have less CaCl2, say 0.5 moles
and we would add it to a solution of Na2SO4 of 1 moles
then all the CaCl2 will all react and there will have been a complete reaction
______________________________ Yes, that is what my intuition (based on that definition) would be. What skeeved me out was the excess sodium. So 1M of Cl-ions would give 0.5M of CaCl2
but if there is also mentioned 1M of Ca-ions Yeah, and 0.5M of excess Ca ions... then I don't see how CaCl2 can be added to the solution ...
because the #M ratio of CaCl2 is 1M Ca and 2M Cl2 so I don't see how
CaCl2 can be added to the solution
This means that no CaCl2 would be added, but another compound(s) 3:34 PM
The only thing I can think of would be CaCl^+ (aq) could that exist?
then we have the problem of it having no extra Cl- counterion
So I don't think that we can use CaCl^+
//I have to go offline now, but I'll try to find a solution for the next time that I go online! Okay, thanks for all the help so far :)