Question

Co(H2O)62+(aq) + 4Cl-(aq) ⇌ CoCl42-(aq) + 6H2O(l)

The cobalt hydrate on the left side of the equation is a pink color. The cobalt chloride on the right side of the equation is a blue purple color.

Q: To what side of the equation did the equilibrium shift when hydrochloric acid was added? To what side of the equation did the equilibrium shift when water was added? How do these substances added relate to the chemical equation?

Answer 1

We have Co(H2O)62+(aq) + 4Cl-(aq) ⇌ CoCl42-(aq) + 6H2O(l)

Kc = [CoCl42-][H2O]⁶ / [Co(H2O)62+][Cl-]⁴

when we add HCl it will diassociate as

HCl (aq) -----> H+ (aq) + Cl- (aq)

this willl increase the Cl- concentraion therefore Q will decrease

Q is nothing but similar to Kc at new concentration

so Q < Kc means the equilibrium will shift towards products side i.e more products will be formed

when we add water the H2O concentraion will increase so Q will decrease as H2O is in numerator for Q

so Q < Kc means the equilibrium will shift towards reactant side i.e more reactants will be formed