# Influence of adding an acid to an equilibrium reaction that is not part of an equilibrium reaction [closed]

Question:

The equation of chemical equilibrium reaction is the following: $$\ce{[Cu(OH2)4]^2+ + 4Cl- <=> [CuCl4]^2- + 4H2O}$$ The $\ce{[Cu(OH2)4]^2+}$ ions are of blue colour, while the $\ce{[CuCl4]^2-}$ ions are of yellow colour. The mixture at equilibrium is of green colour (meaning that there are both types of ions inside it). How does the colour of the mixture change, if $\ce{HCl(aq)}$ is added?

I know that, if a reactant is added to the equilibrium reaction, the chemical equilibrium is "moved" towards the products, vice versa.

However, the hydrochloric acid $\ce{HCl(aq)}$ is not a part of the equilibrium equation, so I don't know how it could in any way influence the chemical equilibrium. I need explanation for this problem.

## closed as off-topic by Mithoron, airhuff, Todd Minehardt, Tyberius, Jon CusterFeb 27 '18 at 15:04

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• HCl splits into H+ and Cl- in solution. – Tyberius Feb 26 '18 at 17:31

## 1 Answer

The equation as written has two typos: one is the lack of a + sign on the left and the other is H20 instead of H2O on the right. They distract from seeing the beauty and intricacy of the problem.

You have cupric ion (hydrated) plus four chloride ions on the left in equilibrium with tetrachlorocupric dianion and water on the right.

The key concept is that the addition of hydrochloric acid is equivalent to adding hydrogen ions and chloride ions. In a way, adding "HCl" is a mental typo, because it disguises the fact that you are really adding H+ and Cl- ions.

The addition of more chloride ions forces the equilibrium to the right.

And thanks to user32369 for the suggestion to edit.