What is the net ionic equation for the precipitation reaction between copper 2 chloride and sodium hydroxide?

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Copper chloride (CuCl2) and sodium hydroxide (NaOH) react with each other and gives blue colour copper hydroxide (Ni(OH)2) precipitate and sodium chloride (NaCl) as the precipitate. Copper chloride is a blue colour solution and sodium chloride is a colourless aqueous solution.

Copper chloride (CuCl2) and sodium hydroxide (NaOH) react with each other and gives blue colour copper hydroxide (Ni(OH)2) precipitate and sodium chloride (NaCl) as the precipitate. Copper chloride is a blue colour solution and sodium chloride is a colourless aqueous solution.

What is the net ionic equation for cuso4 and NaOH?

The net ionic equation for the precipitation of copper(II) carbonate is Cu+2(aq)+COu221223(aq)u27f6CuCO3(s) C u + 2 ( a q ) + C O 3 u2212 2 ( a q ) u27f6 C u C O 3 ( s ) .

What is the net ionic equation for NaOH and CuSO4?

How to Balance: CuSO4 + NaOH Cu(OH)2 + Na2SO4| Breslyn.org.

What does CuSO4 and NaOH make?

Given- The chemical reaction equation is CuSO4 (aq)+NH4Cl (aq) CuS O 4 ( aq ) + N H 4 Cl ( aq ) . Thus, the complete and net ionic equation is Cu+ (aq)+Clu2212 (aq)CuCl (s) C u + ( aq ) + C l u2212 ( aq ) CuCl ( s ) .

What is the net ionic equation for CuSO4 NaOH?

How to Balance: CuSO4 + NaOH Cu(OH)2 + Na2SO4| Breslyn.org.

What is the ionic equation of CuSO4?

Given- The chemical reaction equation is CuSO4 (aq)+NH4Cl (aq) CuS O 4 ( aq ) + N H 4 Cl ( aq ) . Thus, the complete and net ionic equation is Cu+ (aq)+Clu2212 (aq)CuCl (s) C u + ( aq ) + C l u2212 ( aq ) CuCl ( s ) .

You're dealing with a double replacement reaction in which two soluble ionic compounds react to form an insoluble solid that precipitates out of the aqueous solution.

In your case, sodium hydroxide, #"NaOH"#, and copper(II) nitrate, #"Cu"("NO"_3)_2#, will dissociate completely in aqueous solution to form cations and anions

#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#

#"Cu"("NO"_3)_text(2(aq]) -> "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#

The reaction will produce copper(II) hydroxide, #"Cu"("OH")_2#, an insoluble ionic compound that precipitates out of solution, and aqueous sodium nitrate, #"NaNO"_3#, another soluble ionic compound.

The balanced chemical equation for this reaction would look like this

#2"NaOH"_text((aq]) + "Cu"("NO"_3)_text(2(aq]) -> "Cu"("OH")_text(2(s]) darr + 2"NaNO"_text(3(aq])#

Now, notice that you need #2# moles of sodium hydroxide for every #1# mole of copper(II) nitrate that takes part in the reaction.

To get the complete ionic equation, rewrite the soluble ionic compounds as cations and anions

#2 xx overbrace(("Na"_text((aq])^(+) + "OH"_text((aq])^(-)))^(color(purple)("NaOH")) + overbrace(("Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)))^(color(red)("Cu"("NO"_3)_2)) -> "Cu"("OH")_text(2(s])# #darr# #+ 2 xx underbrace(("Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)))_color(blue)("NaNO"_3)#

This is equivalent to

#2"Na"_text((aq])^(+) + 2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-) -> "Cu"("OH")_text(2(s]) darr + 2"Na"_text((aq])^(+) + 2"NO"_text(3(aq])^(-)#

Now, in order to get the net ionic equation, you must eliminate spectator ions, i.e. ions that are present on both sides of the equation

In this case, you would have

#color(red)(cancel(color(black)(2"Na"_text((aq])^(+)))) + 2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-)))) -> "Cu"("OH")_text(2(s]) darr + color(red)(cancel(color(black)(2"Na"_text((aq])^(+)))) + color(red)(cancel(color(black)(2"NO"_text(3(aq])^(-))))#

which is equivalent to

#2"OH"_text((aq])^(-) + "Cu"_text((aq])^(2+) -> "Cu"("OH")_text(2(s]) darr#

Copper(II) hydroxide is a blue insoluble solid that precipitates out of solution.

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