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NaOH + CuSO4


NATT

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Bros i wanna know what will be the the result. I added NaOH with water and hope i made it saturated cuz i mixed em till nothing more dissolved in it. Then i put a piece of CuSO4 and they reacted very rapidly and the solution became blue and i saw a gas was emmited. Any Ideas?????

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The blue solution is Na2SO4???

 

No the blue precipitate is [math]Cu(OH)_{2\ s}[/math]

 

The overall reaction is:

 

[math]CuSO_{4\ aq} + 2OH^{-}_{aq} \rightarrow Cu(OH)_{2\ s} + SO^{2-}_{4\ s} [/math]

 

The blue in the solution is caused by the unreacted [math]Cu^{2+}_{aq}[/math]

Edited by DJBruce
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I saw some kinda solid particals dipsoted in the bottom.... Are they Na... something


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No the blue precipitate is [math]Cu(OH)_{2\ s}[/math]

 

The overall reaction is:

 

[math]CuSO_{4\ aq} + 2OH^{-}_{aq} \rightarrow Cu(OH)_{2\ s} + SO^{2-}_{4\ s} [/math]

 

The blue in the solution is caused by the unreacted [math]Cu^{2+}_{aq}[/math]

 

what do you think about the gas emitted?

Could u plz temme the whole reaction.......

 

what will happend to Na in NaOH

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I saw some kinda solid particals dipsoted in the bottom.... Are they Na... something

 

As you said you saturated the solution with [math]NaOH_{s}[/math] it could be. If the temperature changed or you added slightly to much [math]NaOH_{s}[/math] then some of it would not dissolve and would simply collect on the bottom. Of course in the above mentioned situation the stuff on the bottom would be [math] NaOH_{s} [/math] not [math] Na [/math].

 

As a small side note their is actually no [math] Na [/math] in the solution their is [math] Na^{+}_{aq} [/math]. Although it seems trivial if there was actually [math] Na_{s} [/math] in the solution you would have a very violent reaction where [math]NaOH_{aq}[/math] is formed


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what do you think about the gas emitted?

Could u plz temme the whole reaction.......

 

what will happend to Na in NaOH

 

I left out the [math] Na^{+} [/math] in the reaction because it is a spectator ion, it simply stays in solution unreacted. If you want the reaction in it entirety here it is:

 

This is the dissociation of the sodium hydroxide.

[math] NaOH_{aq}\rightarrow Na^{+}_{aq} + OH^{-}_{aq}[/math]

 

This is the dissociation of the Copper (II) Sulfate.

[math] CuSO_{4} \rightarrow Cu^{2+}_{aq} + SO^{2-}_{aq} [/math]

 

When you mix them you get this overall reaction.

[math] 2Na^{+}_{aq} + 2OH^{-}_{aq} + Cu^{2+}_{aq} + SO^{2-}_{aq} \rightarrow Cu(OH)_{2 s} + SO^{2-}_{4 aq} + 2Na^{+}_{aq}[/math]

Edited by DJBruce
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As you said you saturated the solution with [math]NaOH_{s}[/math] it could be. If the temperature changed or you added slightly to much [math]NaOH_{s}[/math] then some of it would not dissolve and would simply collect on the bottom. Of course in the above mentioned situation the stuff on the bottom would be [math] NaOH_{s} [/math] not [math] Na [/math].

 

As a small side note their is actually no [math] Na [/math] in the solution their is [math] Na^{+}_{aq} [/math]. Although it seems trivial if there was actually [math] Na_{s} [/math] in the solution you would have a very violent reaction where [math]NaOH_{aq}[/math] is formed


Merged post follows:

Consecutive posts merged

[math]\rightarrow[/math]

 

I left out the [math] Na^{+} [/math] in the reaction becuase it is a spectator ion. If you want the reaction in it entirity here it is:

 

This is the disasociation of the sodium hydoxide.

[math] NaOH_{aq}\rightarrow Na^{+}_{aq} + OH^{-}_{aq}[/math]

 

This is the disacstion of the Copper (II) Sulfate.

[math] CuSO_{4} \rightarrow Cu^{2+}_{aq} + SO^{2-}_{aq} [/math]

 

When you mix them you get this overall reaction.

[math] 2Na^{+}_{aq} + 2OH^{-}_{aq} + Cu^{2+}_{aq} + SO^{2-}_{aq} \rightarrow Cu(OH)_{2 s} + SO^{2-}_{4 aq} + 2Na^{+}_{aq}[/math]

 

Then cant Na2SO4 or kind of that thing will produce by reacting 2Na+ and SO4 2-

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What part of the last equation don't you understand? On top of that, sodium sulphate is a fairly soluble salt, ergo it does not form a precipitate that quickly, considering copper sulphate is far less soluble and you will not have much sulphate in the first place.

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  • 4 months later...

Dear NATT!

 

I can't give you an exact answer, but I wish I could join your experiment! :) We know so little about the concentrated solutions... Here pH is high, and CO2 dissolves easily: OH- + CO2 <-> HCO3- etc. So, it can be CO2, cause you disturbe the system. But it is more funny, if you put a smoldering stick where the gas leaves, and it begins to shining. The emitted gas can be oxygen! Copper is stable in Cu+ form in alkalines, cause the solution creates reductive environment. And the reaction between Cu2+ and some oxygen(II)-containing components can happen, by the reduction of the Cu2+ ion. And it would give you Cu+ and O2. Please examine the emitted gas! I'm curious... I will do it myself in 10 day's time...

 

happy new year, NATT!

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