What is more acidic from [CuCl4]^(-2) and ethanoic acid? And what is the reason?
My gut feeling is the acetic acid is more acidic but I couldn't prove that.
Would [CuCl4]^(-2) be hydrolyzed? :
Cl- <===> HCl + OH-
But Cl^(-) is a strong acid anion..
Cl^- is such a weak base that it does not hydrolyze. Think NaCl in aqueous solution. Neither Na^+ nor Cl^- hydrolyzes so NaCl solutions are neutral. That's unlike NH4Cl where the NH4^+ hydrolyzes (Cl^- does not) to make an acidic solution or sodium acetate (NaAc) where the Ac^- hydrolyzes (Na^+ does not) to make a basic solution. Your last sentence says it all; i.e., Cl^- is a strong acid anion. Therefore, it MUST be a weak base. It can't pull that H^+ away from HOH.
Thank you for the explanation!
To determine which compound is more acidic between [CuCl4]^(-2) (tetrachlorocuprate(II) ion) and ethanoic acid, you need to consider the factors that affect acidity.
1. Presence of hydrogen (H) ions: Acidity is related to the presence of hydrogen ions that can be donated. In ethanoic acid, CH3COOH, there is a hydrogen atom bonded to the carboxyl group (-COOH), which can be easily dissociated as H+ ions. [CuCl4]^(-2) does not contain any hydrogen ions, so it cannot donate H+ ions.
2. Stability of the conjugate base: Acids donate hydrogen ions, which form their conjugate bases when dissociated. The stability of the conjugate base affects the acidity of the acid. The more stable the conjugate base, the stronger the acid. In the case of ethanoic acid, the acetate ion (CH3COO^-) is formed as the conjugate base, which is relatively stable due to resonance stabilization.
On the other hand, [CuCl4]^(-2) forms the tetra-chloro cuprate complex ion as its conjugate base, which is less stable compared to acetate ion.
Therefore, ethanoic acid is more acidic than [CuCl4]^(-2) because it can easily donate H+ ions, and its conjugate base is more stable.