# why do we consider CuO as a base?

According to Arrhenius acid/base theory, a base is a substance which when dissolved in water furnishes $\ce{OH-}$.

But $\ce{CuO}$ is insoluble in water and does not produce $\ce{OH-}$, so why do we consider it a base?

There are two other widely used acid/base theories that apply here: Brønsted–Lowry theory and Lewis theory.

A Brønsted base is defined as a proton acceptor. Technically, $\ce{CuO}$ is capable of reacting with $\ce{H+}$ to form hydroxide ions and eventually $\ce{H2O}$. Although $\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\ce{CuO}$ is present as an undissolved solid).

A Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\ce{CuO}$ has lone pairs of electrons available.

$\ce{CuO}$ is considered a base because it fits these two descriptions.

• Though in practice making bulk $\ce{CuO}$ act as a base is not an easy feat. – permeakra Feb 14 '18 at 6:45

An additional way to consider it: $\ce{CuO}$ is defined as a base in the Lux-Flood theory because it is an oxide ion donor; it donates oxide to oxide acceptors (Lux-Flood acids)

An example of a Lux-Flood acid-base reaction involving $\ce{CuO}$ is the high pressure reaction with carbon dioxide to give $\ce{CuCO_3}$ (source).

$\ce{CuO}$ is a basic oxide and when it reacts with water it gives $\ce{Cu(OH)2}$ which is a base as it gives $\ce{Cu^2+}$ and $\ce{OH-}$ ions when dissolved in water. So technically $\ce{CuO}$ is a base rather a basic oxide. Generally all metallic oxides and hydroxides are bases.

• I know you use the word "generally", but not all metal oxides act (mostly) as bases - CrO3 for instance is best considered acidic. And pedant alert, you mean metal oxide, VO is a metallic oxide, Al2O3 isn't. – Ian Bush Feb 14 '18 at 13:34