Using Coulomb's Law to calculate the energy of a cadmium ion and a sulfide ion at their equilibrium ion-pair separation distance. what do you do?

I do not know off hand...

We know that Coulomb's law states that...

#F_"electrostatic"=k_"e"(q_1q_2)/r^2#

...where #k_"e"= 9.0xx10^9*N*m^2*C^-2#...

...we need #r#, the distance between the cadmium and sulfide ions...

You have not given the equilibrium ion pair distance so I will assume the ions are touching and so the separation r will be the sum of the ionic radii:

#sf(r_(Cd^(2+))=0.097color(white)(x)nm)#

#sf(r_(S^(2-))=0.184color(white)(x)nm)#

#:.##sf(r=0.097+0.184=0.281color(white)(x)nm)#

The force in Newtons between 2 charges #sf(q_1)# and #sf(q_2)# is given by Coulomb's Law:

#sf(F=k.(q_1q_2)/(r^2))#

k is a constant with the value #sf(9.0xx10^(9)color(white)(x)Nm^(2)C^(-2).)#

To find the ion pair energy we need to find the work done in separating the ions from a distance r to #sf(oo)#.

We know that work done W = force x distance moved in the direction of the force.

The problem here is that as we separate the ions, the distance increases so the force decreases i.e it is not constant.

This means we need to do a bit of calculus:

#sf(W=int_r^ooF.dr)#

#:.##sf(W=kq_1q_2int_r^oo1/(r^2).dr)#

#sf(W=-kq_1q_2[1/r]_r^oo)#

#sf(W=-k(q_1q_2)/(r))#

Using the electronic charge to be #sf(1.6xx10^(-19)color(white)(x)C)# we get:

#sf(W=(9.0xx10^(9)xx2xx1.6xx10^(-19)xx2xx1.6xx10^(-19))/(0.281xx10^(-9))color(white)(x)J)#

#sf(W=3.3xx10^(-18)color(white)(x)J)#