In solid sodium chloride (table salt), chloride ions have one more electron than they have protons, and sodium ions have one more proton than they have electrons. These ions are separated at about 0.28 nm. Calculate the electrostatic force between the sodium ion and the chloride ion.
What I did was the following:
F=k|q1*q2|/r^2
F=8.99e9(1.602e-19*1.602e-19)/(2.8e-10)^2
F= 2.94e-9 N
But that's not the answer. Can someone please tell me what I'm doing wrong.
I don't see anything wrong with it.
To calculate the electrostatic force between the sodium ion (Na+) and the chloride ion (Cl-), you correctly applied Coulomb's law, which states that the force between two charged particles is proportional to the product of their charges divided by the square of the distance between them.
However, there is a mistake in the calculation. Let's go through the correct steps:
1. Use the value of the proportionality constant, k, which is approximately 8.99 x 10^9 N m^2/C^2. This constant relates the force to the units of charge and distance in the equation.
2. The charges of Na+ and Cl- ions are +1 and -1, respectively. In the equation, q1 and q2 should be the magnitudes of the charges, so they become 1.602 x 10^-19 C for each ion.
3. The distance between the ions, r, is given as 0.28 nm, which is 2.8 x 10^-10 m.
Now let's recalculate the electrostatic force:
F = (8.99 x 10^9 N m^2/C^2) * |(1.602 x 10^-19 C) * (1.602 x 10^-19 C)| / (2.8 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2/C^2) * (2.573 x 10^-38 C^2) / (7.84 x 10^-20 m^2)
F ≈ 2.934 x 10^-9 N
So, the correct electrostatic force between the sodium ion and the chloride ion is approximately 2.934 x 10^-9 N.