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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
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
Rather than reask the question, it's best to start a bounty on the original and ask for a more definitive answer. –  jonsca Aug 12 '13 at 23:51
@jonsca : He doesn't have enough reputation to offer bounty on a question. –  ashu Aug 12 '13 at 23:54
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 '14 at 10:58

2 Answers 2

The salt is added to vinegar to speed up the process.

Addition of NaCl to vinegar shifts the equilibrium of acid dissociation, i.e. $\ce{CH3COOH <=> CH3COO- + H+}$, to right therefore concentration of $\ce{H+}$ & $\ce{CH3COO-}$ increases. Due to this, $\ce{NaCl}$ reacts with $\ce{CH3COOH}$ to give $\ce{HCl}$

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

Cooper Oxide (base) reacts with acids to form water and copper salts.

Acetic acid ($\ce{CH3COOH}$), which is a weak acid, will react slowly with Copper oxide as compared to $\ce{HCl}$ which is strong acid.

So when use vinegar-salt solution to clean penny, Copper Oxide parellely reacts with HCl and CH3COOH both and gives Copper salt and water.

$\ce{CuO + CH3COOH -> 2H20 + Cu(C2H3O2)2}$

$\ce{CuO + 2HCl -> H2O + CuCl2}$

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I'm still not entirely convinced that adding NaCl will shift the equilibrium. It's definitely possible, but I don't see a good argument for why it should happen that way. –  chipbuster Aug 13 '13 at 0:30
@chipbuster, I'd be strongly inclined to agree. Judging by pKa, HCl is around twelve orders of magnitude more acidic than HOAc, so even in a totally saturated solution the quantity of HCl generated would have to be miniscule. Unfortunately, I can only hazard a guess myself as to the reason that addition of salt accelerates the dissolution. If the solution is saturated, then the lower concentration of water itself could possibly drive the equilibrium forward by favoring dissolution and formation of water from the oxide in the process. That's entirely speculative, though. –  Greg E. Aug 13 '13 at 0:51
There is a slight effect due to the Debye-Huckel effect. –  ashu Aug 13 '13 at 0:59
To a first order approximation though, how strong would such an effect be? –  chipbuster Aug 13 '13 at 6:50
still it will show enough effect because HCl is much more acidic than acetic acid. –  ashu Aug 13 '13 at 8:46

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
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
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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