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Calculate the attractive force between a pair of K and Br- ions that just touch each other. Assume the ionic radius of K+ is 0.133nm and that of Br- to be 0.188nm. The answer should be in piconewtons and should have 1 decimal of accuracy.

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Answer:

The answer is 8.0 x 10⁻⁸ pN

Explanation:

The equation of Coulomb's law is:

[tex]F=k_{e} \frac{q_{1}.q_{2} }{d^{2}}[/tex]

whereas:

  • ke, Coulomb's constant, approx. 9 x 10⁹ N.m².C⁻²
  • q₁, q₂: the quantity of charge of object 1 and object 2
  • d: distance between two object, measured in meter

In our case,

  • The absolute value of the charge of both K⁺ and Br⁻ is equal to the charge of an electron or proton, which is approx. 1.60 x 10⁻¹⁹ C
  • d: distance between two object is the sum of both ion radius, which is 0.133 nm + 0.188 nm = 0.321 nm = 0.321 x 10⁻⁹ m

So the result is

9 x 10⁹ x 1.60 x 10⁻¹⁹ x 1.60 x 10⁻¹⁹ / 0.321 x 10⁻⁹ = 8.0 x 10⁻²⁰ N = 8.0 x 10⁻⁸ pN

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