A sodium ion, \(\mathrm{Na}^{+}\), with a charge of $1.6 \times 10^{-19} \mathrm{C}\( and a chloride ion, \)\mathrm{Cl}^{-}\(, with charge of \)-1.6 \times 10^{-19} \mathrm{C}\(, are separated by a distance of \)0.50 \mathrm{nm}$. How much work would be required to increase the separation of the two ions to an infinite distance?

First, we'll convert the given distance between the two ions from nanometers (nm) to meters (m). Given distance = \(0.50 \mathrm{nm}\) 1 nm = \(10^{-9} \mathrm{m}\) Distance in meters = \(0.50 \mathrm{nm} \times 10^{-9} \mathrm{m/nm} = 0.50 \times 10^{-9} \mathrm{m}\)

We can now calculate the electrostatic force between the Sodium and Chloride ions using Coulomb's Law: Coulomb's Law: \(F = k \frac{q_1 q_2}{r^2}\) where: - \(F\) is the electrostatic force, - \(k\) is the electrostatic constant, which is \(8.99 \times 10^{9} \mathrm{N m^2 C^{-2}}\), - \(q_1\) is the charge of the Sodium ion (\(1.6 \times 10^{-19}\mathrm{C}\)), - \(q_2\) is the charge of the Chloride ion (\(-1.6 \times 10^{-19}\mathrm{C}\)), - \(r\) is the distance between the ions which is \(0.50 \times 10^{-9} \mathrm{m}\).

Now that we have the electrostatic force, we can compute the potential energy of the current setup. The potential energy associated with the electrostatic force is given by the following formula: \(PE = k \frac{q_1 q_2}{r}\) We can simply insert the known values into this formula to find the potential energy: \(PE = (8.99 \times 10^{9} \mathrm{N m^2 C^{-2}})(1.6 \times 10^{-19} \mathrm{C})(-1.6 \times 10^{-19} \mathrm{C})/(0.50 \times 10^{-9} \mathrm{m})\)

The work done is equal to the change in potential energy when the distance between the ions is increased to infinity. When the distance between the ions is infinite, the potential energy becomes zero. Therefore, the work done is equal to the difference between the potential energy at infinity and the potential energy at the initial distance: Work done = \(PE_\infty - PE_{initial}\) Since \(PE_\infty = 0\), the work done is simply equal to the negative of the initial potential energy: Work done = \(-PE_{initial}\) We can now substitute the initial potential energy from step 3 to get the work done: Work done = \(-(-8.99 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19} / (0.50 \times 10^{-9}))\)

Now, we just need to compute the final value for the work done: Work done = \(8.99 \times 10^{9} \times (1.6 \times 10^{-19})^2 / (0.50 \times 10^{-9})\) Work done = \(2.87\times 10^{-19} \mathrm{J}\) Thus, the work required to increase the separation between the Sodium ion and the Chloride ion to an infinite distance is approximately \(2.87\times 10^{-19} \mathrm{J}\).