[moeliya]

What kind of reaction will occour and what is the final balanced reaction ?
Cobalt(II) chloride (0.1) + nitric acid ?
Cobalt(II) chloride (0.1 M)+ h2So4 ?
Cobalt(II) chloride (0.1) + HCl ?
Can you guys tell me how to think ? i think the last one will be a complex but am not sure.

[magician4]

this is not something to "think your way through with reasoning", as cobalt(II) / cobalt (III) ions chemistry often is too close to call with respect to standard chemical principles : a lot of what happens there you'll have to learn

so, pic up a good book (Hollemann-Wiberg would be my personal first choice) and read what is known to happen


regards

Ingo

[moeliya]

due to all respect, i have  read but I dont still get it, i know the last one will be blue and its might be a complex but i cant understand more of this. Could you please help me ?

[magician4]

so, let me give you a brief introduction into selected aspects of Co - ions chemistry ...

Co ions will show up in two different "prominent" oxidation states (+ II , + III) , where the respective stability highly depends on the nature of the ligand system accompanying the ion.

this includes a powerfull desire to change oxidation states, shouldn't the ligands at hand be "just right":
(a)  Co²⁺ in presence of NH₃ badly wants to become oxidized, and turn [Co(NH₃)₆]³⁺
4 CoCl₂ + 4 NH₄Cl + 20 NH₃ --( oxygen, cat. carbon)--> 4 [Co(NH₃)₆]Cl₃ + 2 H₂O
(b)  Co₂(SO₄)₃   in presence of water however will aggressively seek to become + 2 again, and even is powerfull enough in this desire to destroy water oxidatively:
2 Co₂(SO₄)₃ + 2 H₂O --(water)--> 4 Co²⁺_(aq.) (plus other complexes including sulfate: this depends on conc.) + 4 H⁺ + O₂ + 6 SO₄^2-

as a rule of thumb, aminoligands and thatlike will stabilize the ox. state of +III  ( but you'll have to learn those, as there's really no clear system ) , whereas "weak" ligands like water , chloride, nitrate, sulfate will stabilize the (+II) oxidation state

so, first thing to check whenever judging possible reactions of cobalt-ions is, whether they might want to change oxidation state due to the environment you'll expose them to.
fortunately, with the problems given, this is not the case ( as all ligands are stabilizers of ox. state +II , and that's the one we're starting with anyway in Co(NO₃)₂)
so, we'll have to consider none redox reaction whatsoever

however, cobalt being a block II element, its ions are suspects for complex formation (this is a second option of what might happen, even with no redox processes involved)
... and in fact that's what we're about to observe here

In particular, the problem given to you seems to me to be meant to teach you which ligands here are "stronger" (and hence will form, replacing other ligands should the occasion arise), and which are not

so , pls. let's take a look at a 0.1 M solution of Co(NO₃)₂ in water first:

Co((NO₃)₂ + 6 H₂O --(water)--> [Co(H₂O)₆]²⁺_(aq.) + NO₃^-_(aq.)
so, under these conditions ( i.e. approx c(H₂O) = 55.5 mol/L , c(NO₃^-)=0.1 mol/L), nitrate is not strong enough to force the replacement of water in the octahedral coordination sphere of the Co²⁺ ion

as , however, this occupation with ligands for a given cation is subject to (sometimes a multitude of)  equilibrium reactions, for example
 [Me(L₁)_x]ⁿ⁺ + y L₂  [Me(L₁)_(x-z)(L₂)_z]ⁿ⁺ + (y-z) L₂ + z L₁
we might wish to test for higher conc. of nitrate, and whether those might shift the [Co(H₂O)₆]²⁺ situation to something like a [Co(H2O)₅(NO₃)]⁺ situation

the result however will be, that nothing happens : nitrate is unable to replace water as a ligand at a Co²⁺ , even at high nitrate concentrations:
[Co(H₂O)₆]²⁺ + HNO₃  no reaction

with chloride, however, the situation is quite different
[Co(H₂O)₆]²⁺ , Cl^- (excess)  [Co(Cl)₄]^2- , water
this change is accompanied by a significant colour change "pale pink / deep blue"
(pls. also note, that along the way cobalt (II) did change the number (and geometry) of coordinated ligands from 6 , octahedral to 4 , tetrahedral )

..and I leave the formulation of a correct, balanced  equation to you

for sulfate, there is a similar process, but you will need high conc. of sulfate to really achieve something here, as the complex forming constant isn't that strong:

[Co(H₂O)₆]²⁺ , SO₄^2- (huge excess)  [Co(SO₄)₂]^2-
( I couldn't find that much more information with respect to this complex in a hurry,  except that it forms with a K-value of approx 2.5
So, if your "chemical heart" is yelling for more information, whether it's of tetrahedral geometry or other, what the colour might be and whether there'd be two more waterligands still present and so on and so forth, you'd have to do some more research on this on your own)

again, I'll leave the formulation of a correct, balanced  equation to you


hope this helps a bit


regards

Ingo

[moeliya]

Thansk for the information sir! But why do you use coIIchloride with water insted of with chloride as my question was ? is it the same thing ?
Is the second and third complex ?
I get that the first equation doesnt work, but the second one and the third one will work.
s the balanced equation for COCl2 )+ h2So4 --> CoSO4(s)   ? and the third one is : 2 HCl + CoCl 2==> [CoCl4]2- + 2H30+

[magician4]

Thansk for the information sir! But why do you use coIIchloride with water insted of with chloride as my question was ? is it the same thing ?

It's simply been late here, that's all: I goofed insofar, pls. accept my apologies

however,  at 0.1 M it doesn't make a difference at all whether you started from the nitrate or the chloride: you'd have [Co(H₂O)₆]²⁺ for starters

Is the second and third complex ?

in waterbased solutions, cobalt is always complex, as [Co(H₂O)₆]²⁺ is a complex, too ("hexaqua - complex")
you only might find that some (or all) of the water ligands might be replaced, resulting in different complexes

is the balanced equation for COCl2 )+ h2So4 --> CoSO4(s)   ?

no
neither does only one sulfate join the cobalt, nor would the CoSO₄ be an insoluble salt if it did occur, which it doesn't
as I said, [Co(SO₄)₂]^2- has to result from this

and the third one is : 2 HCl + CoCl 2==> [CoCl4]2- + 2H30+

in principle , that's correct
however, as this is to be an equation (i.e. "equal" left to right) , it might be a good idea to have 2 molecules of water listed on the left side, too, if you wish to show H₃O⁺ instead of just H⁺ on the right side


regards

Ingo

[moeliya]

so the second one will be :
 CoCl 2 (aq)   + 2H2SO4(aq)    ->[Co(SO4  ) 2] 2- (s)  +4H  + 2Cl   ?
And the third if I write :
2Cl- +CoCl 2==> [CoCl4]2-  is it still right ? or do i have to elimnate the chloride from the left side ?

[moeliya]

Ive also tried to put the second reaction into http://www.webqc.org/balance.php but it says its impossible so whats wrong with it ? does it even occur

[magician4]

so the second one will be :
 CoCl 2 (aq)   + 2H2SO4(aq)    ->[Co(SO4  ) 2] 2- (s)  +4H + 2Cl   ?

if as a last refinement you'd add the charges at the red marked atoms, it'll be all right

And the third if I write :
2Cl- +CoCl 2==> [CoCl4]2-  is it still right ?

yep, this also is correct
(note: this is "shorthand" each chemist will understand. however, your professor might wish you to show complete substances )

regards

Ingo

[moeliya]

So the secont ne is right with the charges on ?  my teacher want a reaction without ions that doesnt react and the state of matter sign. Sir I dont even know how to thank you!!! so

[moeliya]

What is the state of matter of the reactants and products ?

[Vidya]

reactant and product which are  soluble in water are in aq phase

[moeliya]

Yes i get that but is [CoCl4]2 , CoCl2  ,  HCl aq ??  is H2SO liquid ?   is [Co(SO4  ) 2] 2 solid ?

[moeliya]

I thinkg my second reactin is wrong cause it should be so4) 2- instead os h2so4

[Vidya]

they all are aqueous because they all are present in water and nothing has come out as solid