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Let's say you have a solution that is vinegar and it has salt ($\ce{NaCl}$) dissolved in it. Then you place old dull pennies in it. Then you look at it 5 minutes later and the pennies are clean and looking like new again.

Why is the salt necessary for this process and what exactly does it do?

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I do not believe that the salt is necessary, but you should definitely try this at home. Copper(II) oxide is not soluble in water by itself, but Copper(I) oxide is soluble in acid. Vinegar is mostly water but a small percent of it is acetic acid. - You can find that under solubility on wikipedia. –  Leonardo Dec 13 '12 at 3:52
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The salt might increase the conductivity and facilitate a redox process. Or it could be there for the common ion effect. $\ce{CuCl2}$ is much more soluble than $\ce{CuO}$. @MDB Does the solution turn blue? –  Ben Norris Dec 14 '12 at 2:19
    
I'd say the sodium chloride is just a catalyst to speed up the reaction given how dilute vinegar is. It forms CuCl2 in minute quantities which enables the neutralisation to speed up. –  user2617804 Apr 24 at 10:58
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1 Answer 1

Copper(II) oxide is basic and reacts with acids to form a copper salt and water.

A strong acid, like HCl, reacts faster than a weak acid - like acetic acid (which is the main acid component of vinegar).

If one adds NaCl to vinegar, you get HCl:

$$\ce{CH3COOH +NaCl<=>CH3COONa +HCl}$$

Copper(II) chloride dissolves in the water.

So, the salt is added to speed up the process.

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The equation above isn't correct. The sodium chloride is totally ionized, the acetic acid is only partially ionized and the amount of hydrogen ion in solution is thus relatively small. –  Paul J. Gans Dec 16 '12 at 2:48
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Thank you, @PaulJ.Gans Essentially what is required is more H+ in solution. I think that the addition of NaCl might shift the equilibrium of acetic acid dissociation towards the right is this equation: $$\ce{CH3COOH<=>H+ +CH3COO-}$$ so that you get more H+. –  AJO_ Dec 16 '12 at 4:27
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To a first approximation the addition of $\ce{NaCl}$ won't shift the equilibrium at all. There is a slight effect due to the Debye-Huckel effect, but that's mostly negligible. –  Paul J. Gans Dec 17 '12 at 2:03
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