Answer:
The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
Explanation:
Let the charge on the cation be q₁
Also let the charge on the anion be q₂
A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C
An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C
The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m
Energy of attraction = [tex]\frac{Kq_1q_2}{r^2}[/tex]
Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂
Energy of attraction = [tex]\frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}[/tex]
Energy of attraction = 1.231 X 10⁻¹¹ J
Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
