Chemistry, 13.07.2019 05:40, liliauedt

Consider that cuo(s) + h2so4(aq) reacts to give cuso4(aq) and cuo(s) + k2so4does not give a product.

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answered: blackbetty79

Answer : 1) CuO is a basic salt and therefore reacts with sulfuric acid to give copper (II) sulfate and water.

2) CuO is very stable and therefore does not dissociate in water and therefore cannot react with the sulfate ion in potassium sulfate.

In first case, $CuO_{(s)} + H_{2}SO_{4}_{(aq)} ---- CuSO_{4}_{(s)} + H_{2}O_{(g)}$

Wherein when CuO was made to react with potassium sulphate it didn't had any product because CuO was not ionised by water in presence of potassium sulphate.

answered: sewolf1234

The lattice energy of CuO (s) is very high, so it does not dissolve in water to give its ions. But it is a Bronsted–Lowry base thus it can react with the Bronsted–Lowry acid such as sulphuric acid but not with Bronsted–Lowry base ${{\mathbf{K}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}}$ .

Further Explanation:

The definition of acids and bases can be expressed in many ways based on different theories, which are as follows:

• According to Arrhenius theory, acid is defined as the one which produces hydrogen ions in a solution, while the base is defined as the one which produces hydroxide ions in a solution.

• According to Bronsted–Lowry theory, the acid in the reaction donates a proton while a base is one that accepts a proton.

• According to Lewis theory, the acid in the reaction accepts a pair of electrons while a base donates a pair of electrons.

Lattice energy is termed as the amount of energy released when the ions that exist in gaseous state combine to form compound. The lattice energy of a compound is inversely related to the size of the ions present in it.

The size of ${\text{C}}{{\text{u}}^{2+}}$ and ${{\text{O}}^{2-}}$ is small and therefore lattice energy of CuO(s) is very high. Thus it does not dissolve in water to give its ions. But since it is a Bronsted–Lowry base thus, it can accept the hydrogen ions from sulphuric acid and form ${\text{CuS}}{{\text{O}}_{\text{4}}}$ and ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ .

Therefore, the complete reaction is,

${\text{CuO}}\left(s\right)+{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)\to{\text{CuS}}{{\text{O}}_{\text{4}}}\left({aq}\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)$

But ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)$ is not a Bronsted–Lowry acid. Therefore CuO (s) can not react with ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\left({aq}\right)$ in a way as it does with the sulphuric acid.