Video: Calculating the Force between Ions Using Coulomb’s Law

In a salt crystal, the distance between adjacent sodium and chloride ions is 3.42 × 10⁻¹⁰ m. What is the force of attraction between the two singly chargedions?

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Video Transcript

In a salt crystal, the distance between adjacent sodium and chloride ions is 3.42 times 10 to the negative 10th metres. What is the force of attraction between the two singly charged ions?

Considering this question, one way to write out salt is to represent it using the symbols of each of its elements in the periodic table, that is, sodium, which is represented by capital N and a, and then chlorine, which makes up the chloride ion, which is represented as capital Cl. We’re told that, in a salt crystal, the distance separating the sodium and chloride ions, which we can refer to using the letter 𝑑, is given as 3.42 times 10 to the negative 10th metres.

Knowing this, we want to solve for the force of attraction between these two ions. And we’re also told that the ions are charged, in particular singly charged, which we’ll talk about in a moment. The fact that we’re solving for the force between two charged particles gives us a clue about what mathematical relationship we’ll use to do this.

Coulomb’s law is a law for the electrical force of attraction or repulsion between two charged particles with charges 𝑄 one and 𝑄 two. The law says that if we take the product of these two charges and then multiply them by something called Coulomb’s constant, represented with a lowercase 𝑘, and then divide all that by the distance between the charges squared, then, in that case, we’ve solved for the electrical force between the charges.

In our case, we can write that electric force, 𝐹 sub 𝑒, which by the way is the force of attraction we want to solve for, as 𝑘 times 𝑄 one, the charge of our sodium ion, multiplied by 𝑄 two, the charge of our chloride ion, all divided by the distance 𝑑 between the ions squared. As we said, 𝑘 Coulomb’s constant is indeed a constant. And the value we’ll use for it is 8.99 times 10 to the ninth newton metres squared per coulomb squared. Those are some complicated looking units. But we’ll see why they are what they are in a moment.

Next, let’s consider the two charges of our ions, 𝑄 one and 𝑄 two. In the problem statement, we’re told that the ions are singly charged. That means that the net charge on each of our two ions, the sodium and chloride ions, is equal to one unit of the smallest possible amount of electric charge, which is the charge of an electron or the charge of a proton. Those charges have opposite signs but the same magnitude, which forms our minimum value for charge.

And what is that minimum value? We can call it 𝑄. And it’s equal to 1.6 times 10 to the negative 19th coulombs. That’s the charge of a proton. Or we could also call it the magnitude of the charge of an electron. The fact that these two ions are singly charged means that each of them has a net charge equal to lowercase 𝑞. That is, 𝑄 one and 𝑄 two are both equal to 1.6 times 10 to the negative 19th coulombs.

Knowing all that, we are now ready to plug in values to solve for 𝐹 sub 𝑒, the force of attraction between these ions. When we insert these values, take a look for a moment at the units. We mentioned how strange the units of Coulomb’s constant 𝑘 appear. But we see now that they make sense. In the context of Coulomb’s law, the units of coulombs cancel out as do the units of metres squared. This means that, thanks to the units of the constant 𝑘, our overall outcome will have units of newtons, the units of force.

And entering this equation on our calculator, we find a result of 1.97 times 10 to the negative ninth newtons. That is equal to the force of attraction between the two singly charged ions.

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