The force of attraction between a divalent cation and a divalent anion is 1.49 x 10-8 n. if the ionic radius of the cation is 0.096 nm, what is the anion radius?

The attractive force between two isolated ions is given by the formula:

 -(1)

where,

is force of attraction, is permittivity of vacuum = , and are valence of two ions, is the electron charge = and is the inter-atomic distance.

The inter-atomic distance,

where, is the radius of cation and is the radius of anion.

Since value of radius of cation is given so,

  -(2)

The cation and anion are divalent so the value of and is 2 and -2 of cation and anion respectively.

Substituting the values in formula (1):

=

Substituting this value of in equation 2:

Hence, the radius of anion is .

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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