based upon an experiment, a equipment that had $ceCu^2+$ created a precipitate with $ceNaOH$. Is this exactly how you compose the equation?

$$ce2NaOH + Cu^2+ -> Cu(OH)2 + 2Na+$$

On the experiment: equipment A is combined with NaOH (solution turned white). Using a fire test equipment A is shade green. Ns figured solution A has actually $ceCu^2+$ ion.

You are watching: Copper(ii) sulfate and sodium hydroxide net ionic equation

Yes, the reaction is correct yet it might be preferable come either encompass all or no spectator ions.Use one of two people a network ionic equations (omit the Na+), molecule equation (include the copper link full) or finish ionic equation.

For example, if the compound is copper sulphate the equations would certainly be (without states)

Net ionic$$ceCu^2++ce2OH^-->Cu(OH)_2$$Molecular equation$$ceCuSO_4+ce2NaOH->Cu(OH)_2+ceNa_2SO_4$$Complete Ionic Equation$$ceCu^2+ +SO_4^2-+ce2Na^+ +2OH^-->Cu(OH)_2+ce2Na^+ +SO_4^2-$$

Edit

In the case where friend don"t know what copper link it is, use a net ionic equation. As a side note, copper hydroxide is not white. You can want to examine your calculations.

If you"re wondering what it is, check what various other compounds will emit green.

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edited Dec 19 "16 at 1:08
answered Dec 19 "16 in ~ 1:03

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